Math of relativistic mass different from that of rest mass ?

[koti]

Why is the math used to describe relativistic mass different from the math used to describe rest mass? As far as I know, the stress tensors look differently in those two cases? Is it the case that gravity behaves differently for relativistic mass and rest mass? I suspect that it cannot be the case but why are the stress tensors different? I find it absolutely stunning that there might be a discrepancy between how gravity behaves for relativistic and rest mass but that's just not possible right ?

[Tim88]

It seems to me that your question is not right...

Perhaps you confuse differences in conventions (definitions) with differences in mathematics and physics.

Can you give an example for matter in which (γm) differs from γm, or in which m is unequal to (γm)/γ?

[koti]

There is a stament here:
https://www.quora.com/Relativity-physics-Does-relativistic-mass-have-gravity
that got me thinking:
"Therefore the stress-energy tensor for a highly relativistic particle does not look like the tensor for a stationary particle with a rest mass equivalent to the relativistic energy of the highly relativistic particle"
If the stress tensors are different, would it be accurate to say that the gravity of a star would have a different value than that of say 1kg accelerated to adequate velocity required for that 1kg to obtain a relativistic mass of that star ?

Edit: in simple lame words: is gravity different for rest mass and for relativistic mass? And if yes, why?
I find this absolutely fascinating.

Edited May 1, 2017 by koti

[Mordred]

You don't need the stress tensor to see the difference..

 

[latex] e=m_oc^2 [/latex] rest mass no monentum term.

 

[latex] e^2=pc^2+(m_oc^2)^2 [/latex]

 

total energy rest mass+momentum. The difference is whether or not your looking at the momentum.

 

Remember gravitational mass is the same as inertial mass as far as the equivalence is involved.

 

Now lets look closer at the field equations. In particular your metric tensor vs the stress tensor.

 

The stress tensor tells space how to curve but is spacetime that tells how individual particles move.

 

[latex] G_{\mu\nu}=\eta_{\mu\nu}+T_{\mu\nu}[/latex]

 

it is the metric tensor that tells an individual how to move via its geodesic equation.

 

However the stress tensor for the localized area tells space how to curve from the Euclidean metric [latex] \eta_{\mu\nu}[/latex]

Edited May 1, 2017 by Mordred

[studiot]

Again short and to the point Mordred. +1

[koti]

You don't need the stress tensor to see the difference..

 

[latex] e=m_oc^2 [/latex] rest mass no monentum term.

 

[latex] e^2=pc^2+(m_oc^2)^2 [/latex]

 

total energy rest mass+momentum. The difference is whether or not your looking at the momentum.

 

Remember gravitational mass is the same as inertial mass as far as the equivalence is involved.

So the momentum is the difference but gravitational mass is the same as inertial mass? Thats a contradiction. I'm sure your post is great like Studiot says but Im just too dense and I don't get it.

[Mordred]

Your thinking in terms of strictly the Lorentz transforms. GR doesn't quite work the same.

 

In the Lorentz transforms the metric is Euclidean flat.

 

[latex]G_{\mu\nu}= \eta_{\mu\nu}+h_{\mu\nu}[/latex] in this instance the stress/momentum term is not required as spacetime itself is Euclidean flat. The h tensor is you individual particle deviation from the Minkoskii tensor [latex] \eta [/latex].

 

one sec going to copy paste to save time.

 

 

Start with the metric

[latex](dx^0)^2-(dx^1)^2-(dx^2)^2-(dx^3)^2[/latex]

 

(please note this equation is a direct translation of Pythagorous theory applied to our coordinates)

 

apply it to the equation of motion of a particle in a gravitational field with geodesic equation

 

[latex]\frac{d^2x^\mu}{ds^2}=\Gamma^\mu_{\lambda\nu}\frac{dx^\lambda}{ds}\frac{dx^\nu}{ds}=0[/latex]

 

[latex] ds^2=-c^2dt^2+dx^2+dy^2+dz^2=\eta_{\mu\nu}dx^{\mu}dx^{\nu}[/latex]

[latex]\eta=\begin{pmatrix}-c^2&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}[/latex]

 

so how does this work with the boosts and rotations.

 

lets first show a boost in the x direction.

 

[latex]\begin{pmatrix}\acute{ct}\\ \acute{x}\\\acute{y}\\\acute{z}\end{pmatrix}[/latex][latex]=\begin{pmatrix}\gamma&-\beta\gamma&0&0\\-\beta\gamma&\gamma&o&0\\0&0&1&0\\0&0&0&1\end{pmatrix}[/latex][latex]\begin{pmatrix}ct\\x\\y\\z\end{pmatrix}[/latex]

 

now a boost in the y direction

[latex]\begin{pmatrix}\acute{ct}\\ \acute{x}\\\acute{y}\\\acute{z}\end{pmatrix}[/latex]][latex]=\begin{pmatrix}\gamma&0&-\beta\gamma&0\\0&0&1&0\\-\beta\gamma&0&\gamma&0\\0&0&0&1\end{pmatrix}[/latex][latex]\begin{pmatrix}ct\\x\\y\\z\end{pmatrix}[/latex]

 

in the z direction

 

[latex]\begin{pmatrix}\acute{ct}\\ \acute{x}\\\acute{y}\\\acute{z}\end{pmatrix}[/latex]][latex]=\begin{pmatrix}\gamma&0&0&-\beta\gamma\\0&1&0&0\\0&0&1&0\\-\beta\gamma&0&0&\gamma\end{pmatrix}[/latex][latex]\begin{pmatrix}ct\\x\\y\\z\end{pmatrix}[/latex]

 

now stop and think about the coordinate ct which gives your length with the appropriate length contraction. remember the inertial frames will not have identical coordinate frames and we need to handle the frame transforms. We use pythagorous theory relations to reflect this in GR.

 

the above three boosts can be compactly written as

 

[latex]\acute{x}=\Lambda(v) x[/latex]

 

the problem gets further complicated when you get acceleration which generates rapidity.

 

[latex]\epsilon^\phi=\gamma(1+\beta=1+v/c[/latex]

 

which gives Lorentz transforms as

[latex]ct-x+\epsilon^{-\phi}(\acute{ct}-\acute{x})[/latex]

[latex]ct=x+\epsilon^{-\phi}(\acute{ct}+\acute{x})[/latex]

[latex]y=\acute{y}[/latex]

[latex]z=\acute{z}[/latex]

 

which is a hyperbolic rotation of your IF frames.

 

[latex]\gamma=cosh\phi=\frac{\epsilon^\phi+\epsilon^{-\phi}}{2}[/latex]

[latex]\beta\gamma=sinh\phi=\frac{\epsilon^\phi-\epsilon^{-phi}}{2}[/latex]

and therefore

[latex]\beta=tanh\phi\frac{\epsilon^\phi-\epsilon^{-\phi}}{\epsilon^\phi-\epsilon^{-\phi}}[/latex]

 

which in matrix form is

 

[latex]\begin{pmatrix}\acute{ct}\\ \acute{x}\\\acute{y}\\\acute{z}\end{pmatrix}[/latex]][latex]=\begin{pmatrix}cosh\phi&-sinh\phi&0&0\\-sinh\phi&cosh\phi&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}[/latex][latex]\begin{pmatrix}ct\\x\\y\\z\end{pmatrix}[/latex]

 

Now the above is your key Lorentz transformations. As our global metric is Euclidean flat we can use the Minkowskii metric.

 

When you have a stress tensor the Minkowskii tensor is no longer used and we now use the metric tensor [latex] G_{\mu\nu} [/latex]

 

Space tells matter how to move

 

Which is what is happening above. Recall that a freefall particle the rate that particle falls is spacetime itself. We don't use force. If you also recall the mass term cancels out in freefall motion.

 

So for the individual particle the mass makes no difference.

 

As far as the equivalence principle itself study the Einstein elevator.

 

https://www.google.ca/url?sa=t&source=web&rct=j&url=http://www.einstein-online.info/spotlights/equivalence_principle&ved=0ahUKEwjCiP-88c7TAhWBKGMKHTHyALEQFghdMA0&usg=AFQjCNHrNM7avF6jzeB_-E7qPRENhHvuRA&sig2=S-AG7TC0e-9pImciIrpyeQ

 

The key detail is

 

[latex] f=m_{inertial}a=m_{gravitational}[/latex] g. g itself is also acceleration therefore

m_i=m_g and the above are equivalent. The Einstein elevator explains why

 

[latex] f=ma [/latex]

[latex] f=mg [/latex]

[latex] m_{i}=m_g [/latex]

 

For now ignore the tidal force though that becomes important in your intrinsic curvature. You will have tidal force when you have curvature not when your in Euclidean flat.

 

the key point is that the stress/momentum tensor is the average of all particles in a region which tells spacetime how to curve.

 

Spacetime itself tells an individual particle how to move. The stress tensor sets your global metric. That global metric tells individual particles how to move not the stress tensor. Hope that helps.

 

Actually lets add a thought statement.

 

Take two particles and let them freefall.

 

If your global metric is Euclidean flat they will fall in a Parallel path to each other. (excusively via principle of equivalence)

 

If your global metric is curved they will have a deviation to their freefall determined by the centre of gravity.(tidal force) Ie both paths will get closer and closer together. (The rate of change is the Kronecker delta function just an FYI).

 

A general rule to follow, when transforming between inertial frames the (at rest frame is your local geometry) the particle motion is the deviations from that reference frame. This includes switching observers. ie one observer at a different gravitational potential.

 

[latex]G_{\mu\nu}= \eta_{\mu\nu}+h_{\mu\nu}[/latex] when you have curvature you would use the stress tensor but it is used to determine your geometry. It is the geometry that determines the freefall path.

Edited May 1, 2017 by Mordred

[koti]

Your thinking in terms of strictly the Lorentz transforms. GR doesn't quite work the same.

 

In the Lorentz transforms the metric is Euclidean flat.

 

[latex]G_{\mu\nu}= \eta_{\mu\nu}+h_{\mu\nu}[/latex] in this instance the stress/momentum term is not required as spacetime itself is Euclidean flat. The h tensor is you individual particle deviation from the Minkoskii tensor [latex] \eta [/latex].

 

one sec going to copy paste to save time.

 

 

Start with the metric

[latex](dx^0)^2-(dx^1)^2-(dx^2)^2-(dx^3)^2[/latex]

 

(please note this equation is a direct translation of Pythagorous theory applied to our coordinates)

 

apply it to the equation of motion of a particle in a gravitational field with geodesic equation

 

[latex]\frac{d^2x^\mu}{ds^2}=\Gamma^\mu_{\lambda\nu}\frac{dx^\lambda}{ds}\frac{dx^\nu}{ds}=0[/latex]

 

[latex] ds^2=-c^2dt^2+dx^2+dy^2+dz^2=\eta_{\mu\nu}dx^{\mu}dx^{\nu}[/latex]

[latex]\eta=\begin{pmatrix}-c^2&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}[/latex]

 

so how does this work with the boosts and rotations.

 

lets first show a boost in the x direction.

 

[latex]\begin{pmatrix}\acute{ct}\\ \acute{x}\\\acute{y}\\\acute{z}\end{pmatrix}[/latex][latex]=\begin{pmatrix}\gamma&-\beta\gamma&0&0\\-\beta\gamma&\gamma&o&0\\0&0&1&0\\0&0&0&1\end{pmatrix}[/latex][latex]\begin{pmatrix}ct\\x\\y\\z\end{pmatrix}[/latex]

 

now a boost in the y direction

[latex]\begin{pmatrix}\acute{ct}\\ \acute{x}\\\acute{y}\\\acute{z}\end{pmatrix}[/latex]][latex]=\begin{pmatrix}\gamma&0&-\beta\gamma&0\\0&0&1&0\\-\beta\gamma&0&\gamma&0\\0&0&0&1\end{pmatrix}[/latex][latex]\begin{pmatrix}ct\\x\\y\\z\end{pmatrix}[/latex]

 

in the z direction

 

[latex]\begin{pmatrix}\acute{ct}\\ \acute{x}\\\acute{y}\\\acute{z}\end{pmatrix}[/latex]][latex]=\begin{pmatrix}\gamma&0&0&-\beta\gamma\\0&1&0&0\\0&0&1&0\\-\beta\gamma&0&0&\gamma\end{pmatrix}[/latex][latex]\begin{pmatrix}ct\\x\\y\\z\end{pmatrix}[/latex]

 

now stop and think about the coordinate ct which gives your length with the appropriate length contraction. remember the inertial frames will not have identical coordinate frames and we need to handle the frame transforms. We use pythagorous theory relations to reflect this in GR.

 

the above three boosts can be compactly written as

 

[latex]\acute{x}=\Lambda(v) x[/latex]

 

the problem gets further complicated when you get acceleration which generates rapidity.

 

[latex]\epsilon^\phi=\gamma(1+\beta=1+v/c[/latex]

 

which gives Lorentz transforms as

[latex]ct-x+\epsilon^{-\phi}(\acute{ct}-\acute{x})[/latex]

[latex]ct=x+\epsilon^{-\phi}(\acute{ct}+\acute{x})[/latex]

[latex]y=\acute{y}[/latex]

[latex]z=\acute{z}[/latex]

 

which is a hyperbolic rotation of your IF frames.

 

[latex]\gamma=cosh\phi=\frac{\epsilon^\phi+\epsilon^{-\phi}}{2}[/latex]

[latex]\beta\gamma=sinh\phi=\frac{\epsilon^\phi-\epsilon^{-phi}}{2}[/latex]

and therefore

[latex]\beta=tanh\phi\frac{\epsilon^\phi-\epsilon^{-\phi}}{\epsilon^\phi-\epsilon^{-\phi}}[/latex]

 

which in matrix form is

 

[latex]\begin{pmatrix}\acute{ct}\\ \acute{x}\\\acute{y}\\\acute{z}\end{pmatrix}[/latex]][latex]=\begin{pmatrix}cosh\phi&-sinh\phi&0&0\\-sinh\phi&cosh\phi&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}[/latex][latex]\begin{pmatrix}ct\\x\\y\\z\end{pmatrix}[/latex]

 

Now the above is your key Lorentz transformations. As our global metric is Euclidean flat we can use the Minkowskii metric.

 

When you have a stress tensor the Minkowskii tensor is no longer used and we now use the metric tensor [latex] G_{\mu\nu} [/latex]

 

Space tells matter how to move

 

Which is what is happening above. Recall that a freefall particle the rate that particle falls is spacetime itself. We don't use force. If you also recall the mass term cancels out in freefall motion.

 

So for the individual particle the mass makes no difference.

 

As far as the equivalence principle itself study the Einstein elevator.

 

https://www.google.ca/url?sa=t&source=web&rct=j&url=http://www.einstein-online.info/spotlights/equivalence_principle&ved=0ahUKEwjCiP-88c7TAhWBKGMKHTHyALEQFghdMA0&usg=AFQjCNHrNM7avF6jzeB_-E7qPRENhHvuRA&sig2=S-AG7TC0e-9pImciIrpyeQ

 

The key detail is

 

[latex] f=m_{inertial}a=m_{gravitational}[/latex] g. g itself is also acceleration therefore

m_i=m_g and the above are equivalent. The Einstein elevator explains why

 

[latex] f=ma [/latex]

[latex] f=mg [/latex]

[latex] m_{i}=m_g [/latex]

 

For now ignore the tidal force though that becomes important in your intrinsic curvature. You will have tidal force when you have curvature not when your in Euclidean flat.

 

the key point is that the stress/momentum tensor is the average of all particles in a region which tells spacetime how to curve.

 

Spacetime itself tells an individual particle how to move. The stress tensor sets your global metric. That global metric tells individual particles how to move not the stress tensor. Hope that helps.

 

Actually lets add a thought statement.

 

Take two particles and let them freefall.

 

If your global metric is Euclidean flat they will fall in a Parallel path to each other. (excusively via principle of equivalence)

 

If your global metric is curved they will have a deviation to their freefall determined by the centre of gravity.(tidal force) Ie both paths will get closer and closer together. (The rate of change is the Kronecker delta function just an FYI).

 

A general rule to follow, when transforming between inertial frames the (at rest frame is your local geometry) the particle motion is the deviations from that reference frame. This includes switching observers. ie one observer at a different gravitational potential.

 

[latex]G_{\mu\nu}= \eta_{\mu\nu}+h_{\mu\nu}[/latex] when you have curvature you would use the stress tensor but it is used to determine your geometry. It is the geometry that determines the freefall path.

 

Gosh Mordred...one of these days I promise you I will get my act together and study the math of relativity (and math in general) as it gets very annoying for me to have a gap which prevents me from engaging in a proper dialog with you...again (remember that other long thread on spacetime where I begged you to describe something without math and you weren't all that happy about it) However... today is not the day I'm afraid I just wanted to understand the principle of how rest mass and relativistic mass contribute to curvature of spacetime. What are the differences? Why are there differences if there are any?

[Mordred]

The only difference is how the two add available energy to our spacetime system. *assuming were specifically dealing with a multi particle system.

 

The first two equations apply in this case. In either case your correlating the first two equations in my first post to the effective mass of the system.

 

It doesn't matter if your involving inertial or rest mass both are forms of mass so your spacetime system will be affected. How to mathematically define spacetime will differ simply to derive your metric.

 

[latex]G_{\mu\nu}.[/latex] a system that strictly comprised of static particles will not require the stress/momentum tensor components of the stress tensor. (recall its a matrix) A system of particles in motion will require us to use the stress tensor to derive your metric tensor.

 

However once you have defined your metric this then determines an individual particles motion.

 

Hrrm probably easiest to see on the stress tensor itself.

 

[latex]T_{\mu\nu}=\begin{pmatrix}\rho&0&0&0\\0&p&0&0\\0&0&p&0\\0&0&0&p\end{pmatrix}[/latex]

 

So lets look at [latex] T_{00}[/latex]

 

this is your rest mass energy density.

[latex] T_{01} [/latex] is your pressure term. Particles with momentum generate pressure via an equation of state.

 

[latex] w=\rho/p [/latex]

Edited May 1, 2017 by Mordred

[koti]

The only difference is how the two add available energy to our spacetime system. *assuming were specifically dealing with a multi particle system.

 

The first two equations apply in this case. In either case your correlating the first two equations in my first post to the effective mass of the system.

 

It doesn't matter if your involving inertial or rest mass both are forms of mass so your spacetime system will be affected. How to mathematically define spacetime will differ simply to derive your metric.

 

[latex]G_{\mu\nu}.[/latex] a system that strictly comprised of static particles will not require the stress/momentum tensor components of the stress tensor. (recall its a matrix) A system of particles in motion will require us to use the stress tensor to derive your metric tensor.

 

However once you have defined your metric this then determines an individual particles motion.

 

Hrrm probably easiest to see on the stress tensor itself.

 

[latex]T_{\mu\nu}=\begin{pmatrix}\rho&0&0&0\\0&p&0&0\\0&0&p&0\\0&0&0&p\end{pmatrix}[/latex]

 

So lets look at [latex] T_{00}[/latex]

 

this is your rest mass energy density.

[latex] T_{01} [/latex] is your pressure term. Particles with momentum generate pressure via an equation of state.

 

[latex] w=\rho/p [/latex]

I know Im using hammer and nails for a heart transplant but please bear with me Mordred.

 

Say we have an empty universe with earth in it and a 1kg steel ball accelerating away from it. Lets leave out QM and assume GR and SR to describe whats happening. Lets accelerate the ball to enough velocity that it will curve spacetime with exactly the same gravity as our earth - 1G. Will the inertial mass of the ball at that stage be equal to the rest mass of the earth?

[Mordred]

Yes you can accelerate the ball to have the same mass as the Earth. If we start with a steel ball with the same diameter as the Earth the two spacetime regions will effectively be identical.

Edited May 1, 2017 by Mordred

[koti]

Yes you can accelerate the ball to have the same mass as the Earth. If we start with a steel ball with the same diameter as the Earth the two spacetime regions will effectively be identical.

 

Thanks Mordred. What got me confused is this excerpt:

"Therefore the stress-energy tensor for a highly relativistic particle does not look like the tensor for a stationary particle with a rest mass equivalent to the relativistic energy of the highly relativistic particle. In addition when you are considering the case of a massive highly relativistic particle with an energy,

E, travelling past a stationary particle (or star or black hole) with a rest-mass comparable to E, then the non-linear equations of general relativity are too complicated to solve analytically so not much can be said about the "force" of gravity in this case"

 

from here:

https://www.quora.com/Relativity-physics-Does-relativistic-mass-have-gravity

Edited May 1, 2017 by koti

[Mordred]

Ah yes could have saved a bit of time posting that first but thats ok. The info in this thread is still useful to understand.

[koti]

Ah yes could have saved a bit of time posting that first but thats ok. The info in this thread is still useful to understand.

I did. I posted it in #3

[MigL]

Without going into all the math, Koti...

Assume you're on earth and you weigh yourself to be 200 lbs.

Then we put you on a small planetoid which we speed up to have a relativistic mass equivalent to Earth's mass.

You again weigh yourself.

Do you think you would still weigh 200 lbs ?

[koti]

Without going into all the math, Koti...

Assume you're on earth and you weigh yourself to be 200 lbs.

Then we put you on a small planetoid which we speed up to have a relativistic mass equivalent to Earth's mass.

You again weigh yourself.

Do you think you would still weigh 200 lbs ?

I know now (thanks to Mordred) that I would weigh 200 pounds on the accelerated planetoid.

[MigL]

No. you wouldn't.

You would weigh as much as that planetoid's rest mass dictates, as you are in the same frame.

You and the planetoid's space-time curvature ( gravity if you will ) are moving along with it.

you are in the same frame, so essentially at rest within it.

[koti]

No. you wouldn't.

You would weigh as much as that planetoid's rest mass dictates, as you are in the same frame.

You and the planetoid's space-time curvature ( gravity if you will ) are moving along with it.

you are in the same frame, so essentially at rest within it.

Okay, Im confused. Both frames (planetoid and earth) would have essentially same effect on spacetime curvature around them like Mordred said yet they would affect my mass differently? Lets assume that Im walking on a planetoid which is travelling at say 0.5 c. Edited May 1, 2017 by koti

[Delta1212]

Okay, Im confused. Both frames (planetoid and earth) would have essentially same effect on spacetime curvature around them like Mordred said yet they would affect my mass differently? Lets assume that Im walking on a planetoid which is travelling at say 0.5 c.

Is the planetoid traveling at 0.5c or is it at rest, and it is the Earth that is traveling at 0.5c?

Edited May 1, 2017 by Delta1212

[MigL]

There are NO relativistic effects in the rest frame.

If the earth was at rest, or moving at 0.9c ( with respect to what ? ), you would weigh exactly the same 200 lbs.

[Mordred]

Yes Migl is hitting you with a little trick of observers. It may sound the same as your ball above but in that instance I assumed the observer is remote to the planetoid system. Then applied the equivalence principle. Inertial mass=gravitational mass.

 

However Migl has given a circumstance where you have two effective geometries due to a different observer reference frame.

 

Ok lets try this common question. You see this one on forums often.

 

"If a particle with mass moves close to the speed of light why doesn't it become a blackhole?."

 

The answer is although a distance observer will measure a high inertial mass sufficient to become a BH in regards to its Schwartzchild radius.

 

The invariant (rest mass) does not change.

 

Now to Migl's question. Key note you are in the same frame as the Planet. So you are effectively at rest so you have no inertial mass and neither will the planet as it is treated in its rest frame

in this instance.

Edited May 1, 2017 by Mordred

[koti]

Assume an empty universe with earth, a planetoid travelling at 0,5 c which has an innertial mass of that of earth at that velocity and me with a rest mass of 200 pounds. Planetoid and earth have the same diameter and are far away from each other.

Edited May 1, 2017 by koti

[Mordred]

Assume an empty universe with earth, a planetoid travelling at 0,5 c which has an innertial mass of that of earth at that velocity and me with a rest mass of 200 pounds. Planetoid and earth have the same diameter and are far away from each other.

Define your observers in relation to the planet to properly answer this one

[Delta1212]

Assume an empty universe with earth, a planetoid travelling at 0,5 c which has an innertial mass of that of earth at that velocity and me with a rest mass of 200 pounds. Planetoid and earth have the same diameter and are far away from each other.

Is the planetoid traveling at 0.5c or is the Earth?

[Mordred]

and where is the person being weighted with respect to which observer.?