Simple explanation of relativity

[Digix]

I suggest that special relativity can be explained in very simple way,
current explanation is realy just mathematical facts with no real meaning.

SR: all we need is E=mc.
using this I can explain light speed limit and time speed change, probably there is nothing more that can be explained using SR.
I would like to know if there is any phenomenon that cant be explained with that single formula.

general relativity actualy have no meaning on earth, since we dont have any significant gravity fields here, but stating that space is curved is absurd. because all starting point of all science is that space is straight by definition, and we use various forces to move objects and not to bend space around them.

Of course I dont say that Einstein's teory is wrong. It is correct but do not explain anything why it is so. It needs translation to human language.

[worzel]

You never answered my question last time you said this. How do you explain the twin paradox only using E=mc² (I assume that's what you meant).

[HenrikOlsen]

Of course I dont say that Einstein's teory is wrong. It is correct but do not explain anything why it is so. It needs translation to human language.

Actually not, it needs an understanding of mathematics by the reader, mathematics is already a human language.

[gwiz]

general relativity actualy have no meaning on earth, since we dont have any significant gravity fields here, but stating that space is curved is absurd. because all starting point of all science is that space is straight by definition, and we use various forces to move objects and not to bend space around them.

Do you use a GPS navigation device? If so, you are using a bit of technology that incorporates general relativity. If it didn't, it wouldn't be as accurate.

[Digix]

Do you use a GPS navigation device? If so, you are using a bit of technology that incorporates general relativity. If it didn't, it wouldn't be as accurate.

I do not question acuracy of relativity, I just dont like expression of it.
for example it says gravity is accleration, but which accelartion?
ve have 2 types a=v'(t) and a=f/m so I would be statisfied if it mention not generaly acceleration, but force/mass ratio, there is no difference but ir disables all time travel speculations
and forces you to have mass in your calculations.

Actually not, it needs an understanding of mathematics by the reader, mathematics is already a human language.

mathematic, is wrong language when you simplify something for example can you calculate ratio a/b? seems simple, but what if a=s/x and b=e/x ? and in our case x=0? you dont care about what you dont know, and get wrong sesults another example sqr(a)*sqr(a). there are lots of examples that can be used to get wrong results.

You never answered my question last time you said this. How do you explain the twin paradox only using E=mc2 (I assume that's what you meant).

I think I explained that. but I can do it again.
for simplification of e=mc I will refer to mass and energy as to same thing.
I replace all relative time, size, speed with only one parameter relative energy.
we have 2 twins or clocks. of course they are born having same energies.
so if you accelerate one to light speed its time will slow down because of (energy)mass increase. time speed change will be prportional to rest mass increase.
we can asume that twins are somehow born having relative speeds, and they dont know which one is moving. but that is realy impossible, because they will be not same mass and that means not twins. the one who is standing will have less energy than one who have speed of light. of course the one with more energy will have slower time speed than another.

My suggestion would be to use litttle diferent formula for relativity:
time= (e/cc+m0)/m0 (objects with no energy have no time)(and this time is not same ar we use. we use time speed in ourr life (seconds per clock seconds))
time speed will be time(system1)/time(of system2)
so all relativity will be energy/time based, not space/time based.
of course result must be same same. but it is easer to connect to quantum teory, which is energy/time based as well.

[HenrikOlsen]

for example it says gravity is accleration, but which accelartion?

No, it says there's no way to distinguish gravity and acceleration when measuring in a single point.

ve have 2 types a=v'(t) and a=f/m so I would be statisfied if it mention not generaly acceleration, but force/mass ratio, there is no difference but ir disables all time travel speculations
and forces you to have mass in your calculations.

We have F=ma, we also have F=dp/dt, and p=mv/(1-v²/c²)^½ (which approaches p=mv for low speeds) which means that for low speeds we have a=dv/dt, this is not true for high speeds where we have a=d[v/(1-v²/c²)^½]/dt

mathematic, is wrong language when you simplify something for example can you calculate ratio a/b? seems simple, but what if a=s/x and b=e/x ? and in our case x=0? you dont care about what you dont know, and get wrong sesults another example sqr(a)*sqr(a). there are lots of examples that can be used to get wrong results.

That is only true if you don't use mathematics correctly.
In your example, the correct mathematical "simplification is that "a/b=s/e except where x=0".
For sqr(a)*sqr(a), it is that either we're working with complex numbers in which case "sqr(a)*sqr(a)=a" is always true, or reqal numbers in which case it's "sqr(a)*sqr(a)=a for non-negative a".

I think I explained that. but I can do it again.
for simplification of e=mc I will refer to mass and energy as to same thing.
I replace all relative time, size, speed with only one parameter relative energy.

Even if you really mean e=mc², this simplification is not really valid, since this only applies to particles at rest and approximately for particles at low speeds.
For a moving particle it's really e=mc²/(1-v²/c²)^½

we have 2 twins or clocks. of course they are born having same energies.
so if you accelerate one to light speed its time will slow down because of (energy)mass increase. time speed change will be prportional to rest mass increase.

That part is sort of correct if you meant close to light speed.

we can asume that twins are somehow born having relative speeds, and they dont know which one is moving. but that is realy impossible, because they will be not same mass and that means not twins. the one who is standing will have less energy than one who have speed of light. of course the one with more energy will have slower time speed than another.

The concept of standing still is meaningless, apparently you're missing that if both twins are in nonaccelerating frames, they each see the other as moving fast and thus having slowed down time relative to themselves.
The whole thought experiment is about what happens when they start at the same velocity and one goes away and comes back.
Changing this invalidate any conclusion you might have achieved.

My suggestion would be to use litttle diferent formula for relativity:
time= (e/cc+m0)/m0 (objects with no energy have no time)(and this time is not same ar we use. we use time speed in ourr life (seconds per clock seconds))
time speed will be time(system1)/time(of system2)
so all relativity will be energy/time based, not space/time based.
of course result must be same same. but it is easer to connect to quantum teory, which is energy/time based as well.

This sounds like an interesting concept, but you need to develop the mathematics a lot more before it's possible to see if you have a viable concept.

[Digix]

No, it says there's no way to distinguish gravity and acceleration when measuring in a single point.

maybe so, but usualy I find that people say that gravity isaccleration, what have incorrect meaning.

We have F=ma, we also have F=dp/dt, and p=mv/(1-v²/c²)^½ (which approaches p=mv for low speeds) which means that for low speeds we have a=dv/dt, this is not true for high speeds where we have a=d[v/(1-v²/c²)^½]/dt

well, my point is to know how acceleation is created. since mass can only be accelerated by using some force.

That is only true if you don't use mathematics correctly.
In your example, the correct mathematical "simplification is that "a/b=s/e except where x=0".
For sqr(a)*sqr(a), it is that either we're working with complex numbers in which case "sqr(a)*sqr(a)=a" is always true, or reqal numbers in which case it's "sqr(a)*sqr(a)=a for non-negative a".

if you make 2 separate formulas, or some restrictions, some of them will be lost. expecialy if they have no meaning. realtivity does not allow time trawel, but all time machine inventors refer to relativity. this means that math is realy used in wrong way.

Even if you really mean e=mc², this simplification is not really valid, since this only applies to particles at rest and approximately for particles at low speeds.
For a moving particle it's really e=mc²/(1-v²/c²)^½

no , e=mc² is always valid. m is all mass(not just rest mass) and e is all energy.
your formula is logicaly not very good. better would be
e=(m/(1-v²/c²)^½)c²
first we calculate mass then we get energy of that total mass.

The concept of standing still is meaningless, apparently you're missing that if both twins are in nonaccelerating frames, they each see the other as moving fast and thus having slowed down time relative to themselves.
The whole thought experiment is about what happens when they start at the same velocity and one goes away and comes back.
Changing this invalidate any conclusion you might have achieved.

well, as I said my explanation is enegy/time based so I actualy do not care what form of energy anyone of twins will gain so no need for acceleration. ( you can heat up one of then up to few milion degrees or drop on to neutron star, result will be same)
what they will see if both are in constant speed, depends on dopler shift, not on time speed change. this is unrelated to relativity.
the usual thought experiment about going and coming back, is easy to explain. first we change it a litle:
one twin in some way gains lots of energy and relativistic mass comparable to its rest mass. its internal non relative energies are same. so from our point of view his electrons will be much more heavy that in our matter. since electron energy does not change speed will decrease. in this way all time speed will decrease for everyone who have relativistic energy.

if I understand correctly SR does not explain that at all. this experiment is used only to ilustrate SR.

This sounds like an interesting concept, but you need to develop the mathematics a lot more before it's possible to see if you have a viable concept.

I know, for now it is just concept, but it should match well to quantum teory, which core is e=h*t. so again it is energy/time based. and light speed just connects that energy/time to distance.
For now I would like to be sure that I can explain everything with only e=mc². if there are more experiments to explain besides twin paradox.

[Irishman]

You miss a detail - F = m * a for constant acceleration.

[Jim]

... realtivity does not allow time trawel, but all time machine inventors refer to relativity. this means that math is realy used in wrong way. ...

Which is probably why no one has invented a working time machine.

But, just because the math - or relativity - is used in the wrong way does not mean that it is expressed in the wrong way. There is a big difference between implication and inference.

My suggestion would be to use litttle diferent formula for relativity:
time= (e/cc+m0)/m0 (objects with no energy have no time)...

If I get you right, you are equating energy and mass. Saying that an object has no energy means it has no mass. That gets e=0 and m0 = 0. That doesn't mean the objects have no time; time becomes indeterminant, meaningless, or infinite depending on your viewpoint, but not zero.

[Digix]

You miss a detail - F = m * a for constant acceleration.

why constant? by the way this is actualy whong here right formula is
a=m/f since acceleation is created by force not force by acceration.

[Digix]

Which is probably why no one has invented a working time machine.

as I know there ate plently of time machine inventions discused. and nearly nobody points that relativity does not alow that. Most usual argument is casuality violation.

But, just because the math - or relativity - is used in the wrong way does not mean that it is expressed in the wrong way. There is a big difference between implication and inference.

I dont say it is wrong, but nearly all purpose is to be use for absurd speculations.
this is because relativity does not explain why everything is so. too much postulates. for example why light speed is absolute limit? lots of people do not understand that, so usualy ignore. Snd SR does not give unbreakable explanation.

If I get you right, you are equating energy and mass. Saying that an object has no energy means it has no mass. That gets e=0 and m0 = 0. That doesn't mean the objects have no time; time becomes indeterminant, meaningless, or infinite depending on your viewpoint, but not zero.

mass energy equivalenve is usualy accepted even if many astronomy scientists decide to deny it.
in case e=0 and m=0 (not m0 just m) we have no object, and again usulay it is accepted that empty space does not have time properety.
however here we go into zeropoint generator area so energy will apear from nowhere somehow, since quantum mechanis does not alow to have zero energy .(i cant talk nuch about that teory, but there exsists some hyphotesies about this)

[DrChinese]

I dont say it is wrong, but nearly all purpose is to be use for absurd speculations.
this is because relativity does not explain why everything is so. too much postulates. for example why light speed is absolute limit? lots of people do not understand that, so usualy ignore. Snd SR does not give unbreakable explanation.

Your concept of the purpose of a theory has severe problems.

First, good theories are good because they are useful. Relativity works, and can be used to predict many things that the E=mc^2 formula does not. Specifically, it is essential for day-to-day use in particle accelerators. It is also used in the GPS system as noted above.

Second, it is not the responsibilty of a theory to explain why it works. Why is c a local speed limit? E=mc^2 doesn't explain this, and neither does anything else. It simply is.

[Digix]

Your concept of the purpose of a theory has severe problems.

First, good theories are good because they are useful. Relativity works, and can be used to predict many things that the E=mc^2 formula does not. Specifically, it is essential for day-to-day use in particle accelerators. It is also used in the GPS system as noted above.

Second, it is not the responsibilty of a theory to explain why it works. Why is c a local speed limit? E=mc^2 doesn't explain this, and neither does anything else. It simply is.

Good teory can esplain why it works usualy using some another theory for explanation or maybe it can use itself for that purpose.
E=mc^2 can explain speed limit easily:
because all energy is always mass then if you give some energy to some particle it means you incease its mass. more mass means less speed. in this way you will go intil you reach c speed, when mass incerase is infinite in comparison to rest mass, or all mass is relativistisc only.
Photon rest mass is always 0 so only way to have energy is to have C speed.

Also relativity is little wrong, light speed only limits energy transfer. so it is not absolute limit theoreticaly.

[rahuldandekar]

Digix, if you really believe E = mc² is a starting point for a new derivation of relativity results, you need to give logical arguments and proof.

I suggest that special relativity can be explained in very simple way,
current explanation is realy just mathematical facts with no real meaning.

too much postulates. for example why light speed is absolute limit?

Too many postulates? No... Infact, SR is famous from it's derivation from a small number (2) of simple postulates:
(From http://www.patent-invent.com/electri...ivity_2.html):

1. First postulate (principle of relativity)

The laws of electrodynamics and optics will be valid for all frames in which the laws of mechanics hold good.
Every physical theory should look the same mathematically to every inertial observer.
The laws of physics are independent of location space or time.
2. Second postulate (invariance of c)

The speed of light in vacuum, commonly denoted c, is the same to all inertial observers, is the same in all directions, and does not depend on the velocity of the object emitting the light. When combined with the First Postulate, this Second Postulate is equivalent to stating that light does not require any medium (such as "aether") in which to propagate.

See? Just two postulates... the second one is the less intuitive one, but is confirmed by a large number of experiments. The thing about starting points for a theory is... they aren't explained by another theory, but by experiments.

Good teory can esplain why it works usualy using some another theory for explanation or maybe it can use itself for that purpose.

Wrong... the starting postulates of a good theory are based on physical facts, confirmed by experiments.

You too are starting with two postulates: on is E = mc² and m = m₀/(1 - v²/c²)^0.5.
How do you explain these two postulates on the basis of another theory? How do you define Energy and mass in the first place?

Note: Special Relativity permits gamma*m₀c² and gamma*m*v as the formulas for energy and momentum... but even these are only confirmed by experiments.

One more thing: There are many ways to derive the consequences of SR... one is the two postulates above, another is the invariance of the spacetime interval, still another is the invariance of the inner product of the 4-momentum. Your approach is, I believe, equivalent to the third one.

Do show us the logical derivation from your postulate.

[HenrikOlsen]

Good teory can esplain why it works usualy using some another theory for explanation or maybe it can use itself for that purpose.
E=mc^2 can explain speed limit easily:
because all energy is always mass then if you give some energy to some particle it means you incease its mass. more mass means less speed. in this way you will go intil you reach c speed, when mass incerase is infinite in comparison to rest mass, or all mass is relativistisc only.
Photon rest mass is always 0 so only way to have energy is to have C speed.

Also relativity is little wrong, light speed only limits energy transfer. so it is not absolute limit theoreticaly.

A problem here is that it fails to explain why light moves at the speed of light, with m=0, it should have infinite speed when deriving from your method.

[rahuldandekar]

A problem here is that it fails to explain why light moves at the speed of light, with m=0, it should have infinite speed when deriving from your method.

Actually, restated as E = m₀²c⁴ + p²c², it does. This is equivalent to the invariance of the inner product of the 4-momentum.

[Digix]

You too are starting with two postulates: on is E = mc² and m = m₀/(1 - v²/c²)^0.5.
How do you explain these two postulates on the basis of another theory? How do you define Energy and mass in the first place?

I dont think so, m = m₀/(1-v²/c²)^0.5
is derived from E = mc²
in this case there is single postulate energy and mass are same.
or mass in only possible storage of energy.
we can use grams instead of joules to express any energy.
C is only some constant. which is equal to light speed, but, it explains why light speed is equal to this constant, not why constant is equal to lihgt speed.
I will try to write mathematical proff deriving other realtivity formulas from E = mc²
but the, mathematic is hard since all functions are recursive.
for example kinetic energy E=(mv^2)/2. m is almost same as E, so m is not costant. and we need integrate this function over all v range to get result.

[rahuldandekar]

I dont think so, m = m₀/(1-v²/c²)^0.5
is derived from E = mc²
in this case there is single postulate energy and mass are same.

I don't think it can be derived from that.

You should not just state results, you should also give proofs. Where's the proof?

[Digix]

well, seems the re is no way to awoid mathematic.

I will continue this topic when mathematical proof will be ready.

[Irishman]

Digix said:
because all energy is always mass then if you give some energy to some particle it means you incease its mass. more mass means less speed. in this way you will go intil you reach c speed, when mass incerase is infinite in comparison to rest mass, or all mass is relativistisc only.

There's something drastically wrong here. Why does adding energy/mass to the system mean less speed? You've added to the total, not shifted between the two.

why constant [acceleration]? by the way this is actualy whong here right formula is a=m/f since acceleation is created by force not force by acceration.

Excuse me, I bungled this. I meant constant mass. Oops, carry on.

[Digix]

There's something drastically wrong here. Why does adding energy/mass to the system mean less speed? You've added to the total, not shifted between the two.

if energy is constant(pay atention which energy), more mass means less speed.
for example if we throw heavy brick and light brick using same amount of energy light one will go faster.

I am somehow puzzled how to integrate fuction over itself, to derive relativity formulas from e=mc^2. for integration I usualy use finite element method, to avoid mathematic.
we have e=1/2*mv^2 and e=mc^2 these two should be enough to derive
m = m0/(1-v^2/c^2)^0.5

[HenrikOlsen]

if energy is constant(pay atention which energy)

The thing that is constant is the sum of all types of energy, potential, kinetic, heat (which is also a form of kinetic energy), elastic, chemical etc. etc. and energy as mass(mc²), there's no specific energy form that's constant, only the sum of all forms of energy.

If you add energy to a particle, you've added energy to the system it's part of, and in that case there's no reason for the speed to decrease, since there's no requirement for energy constancy.

[rahuldandekar]

if energy is constant(pay atention which energy), more mass means less speed.
for example if we throw heavy brick and light brick using same amount of energy light one will go faster.

I am somehow puzzled how to integrate fuction over itself, to derive relativity formulas from e=mc^2. for integration I usualy use finite element method, to avoid mathematic.
we have e=1/2*mv^2 and e=mc^2 these two should be enough to derive
m = m0/(1-v^2/c^2)^0.5

Yes, but, remember, you need to explain the relation between energy and speed. There's no formula except the aforementioned m in terms of m_o and v one... and that you haven't done yet.

Also... we don't have E = 0.5mv^2 in relativity, you know. We (you) just have E = mc^2, and unless I can find a proof, I ssume it is independent of m = m_o/(1- v²/c²)^0.5 .

So, you'd have to start with two postulates... E = gamma*m_o*c^2 and P = gamma*m_o*v ... (I think that is equivalent to stating the relativistic mass formula and the energy-mass equation... I may be mistaken here... )

Unless I find (i.e. you state) a proof, I will assume E = m*c^2 is independent of m = m_o*gamma.

[Fortis]

we have e=1/2*mv^2 and e=mc^2 these two should be enough to derive
m = m0/(1-v^2/c^2)^0.5

Except that

e=(1/2)mv² (1)

is an approximation. Any derivation that have that assumes that equation (1) is correct will fail to provide a valid relativistic result.

[Digix]

Except that
e=(1/2)mv² (1)
is an approximation. Any derivation that have that assumes that equation (1) is correct will fail to provide a valid relativistic result.

that is corect, and that is reason why it is not so easy to use this formula here.

fors start if I use
e=(m-m0)C^2 and e=1/2 mv^2 (later i will think how logicaly remove 1/2)

in that case I get m=m₀/(1-v²/2c²)
still missing square root, but almost what we need.

I think using correct kinetic energy formula, should fix that problem
since mass is nonconstant, I suppose it will be in same order as speed. so result will be that missing square root.

[rahuldandekar]

that is corect, and that is reason why it is not so easy to use this formula here.

fors start if I use
e=(m-m0)C^2 and e=1/2 mv^2 (later i will think how logicaly remove 1/2)

in that case I get m=m₀/(1-v²/2c²)
still missing square root, but almost what we need.

I think using correct kinetic energy formula, should fix that problem
since mass is nonconstant, I suppose it will be in same order as speed. so result will be that missing square root.

One missstep... you can only say E = (m-m₀)c² when you know that m is different from m₀ ... which is part of what you have to prove. I am getting confused here... I admit.

Please state your postulate(s) along with the definitions for E, and m and m₀. Let's treat this like a real scientific discussion.

Sorry, I get confused if we don't have the right definitions, etc.

[Digix]

For now I suppose the single postulate is enough: is energy is same as mass.
but there is one point, that energy is always relative to something. so we must keep track which energy we calculate and which mass.
Mass becomes relative, what is not common is physic.
E = (m-m₀)c²
in this formula E is kinetic energy of closed system, so we must separate rest mass (energy) which do not contribute to kinetic energy.

if we assume that patricle have no m₀ then we get easy soliution
e=mC^2 and e=1/2 mv^2 (becasue single particle cant have energy we need 2 particles with oposite moving directions and equal speed to release energy.) so we get that energy of closed system is e=mv^2 without 1/2)
so we get v^2=C^2
in this way we get that particle with no rest mass will have speed equal to C

however ir there is rest mass everything is much more complex.

[Fortis]

that is corect, and that is reason why it is not so easy to use this formula here.

fors start if I use
e=(m-m0)C^2 and e=1/2 mv^2 (later i will think how logicaly remove 1/2)

in that case I get m=m₀/(1-v²/2c²)
still missing square root, but almost what we need.

I think using correct kinetic energy formula, should fix that problem
since mass is nonconstant, I suppose it will be in same order as speed. so result will be that missing square root.

You seem to be placing the cart before the horse here. You can derive an approximate form of an equation from the full form of an equation. What you appear to be attempting is to derive the full form from the approximation.

Something that may help you understand what is going on here. Starting with the relativisitic equation for kinetic energy.

KE = m₀c²(gamma-1)

where gamma = 1/(1-v²/c²)^(1/2)

for small v, i.e. v<<c, we can approximate gamma by

gamma =approx 1+(1/2)v²/c²

if we substitute that back into our original expression for KE, we get the classical Newtonian expression (that you use as a starting point)

KE =approx (1/2)m₀v²

What you are attempting to do is to go the other way. Sadly there are an infinite number of functions that would give the classical result as an approximation, and you do not have any means to choose between them.

[Digix]

You seem to be placing the cart before the horse here. You can derive an approximate form of an equation from the full form of an equation. What you appear to be attempting is to derive the full form from the approximation.

Something that may help you understand what is going on here. Starting with the relativisitic equation for kinetic energy.

KE = m₀c²(gamma-1)

where gamma = 1/(1-v²/c²)^(1/2)

for small v, i.e. v<<c, we can approximate gamma by

gamma =approx 1+(1/2)v²/c²

if we substitute that back into our original expression for KE, we get the classical Newtonian expression (that you use as a starting point)

KE =approx (1/2)m₀v²

What you are attempting to do is to go the other way. Sadly there are an infinite number of functions that would give the classical result as an approximation, and you do not have any means to choose between them.

this is not related to my proof, I am not going to derive kinetic energy formula.
but I will choose integral form of kinetic energy formula for deriving gamma.
in newtoniain physic m/m₀ ratio is small so we can asume that m is constant and only integrate speed.

[worzel]

in newtoniain physic m/m₀ ratio is small so we can asume that m is constant and only integrate speed.

I thought that ratio was always exactly 1 in Newtonian physics.