relativity question

[foofighter]

the idea that as an object approaches light speed, its mass increases, seems to contradict the law of conservation of mass. if an object of given mass is speeding up in a vacuum, why should more mass suddenly exist simply because its approaching light speed? where does this extra mass come from? thanks

[insane_alien]

E=mc^2

 

mass is energy and energy is mass. if you gain energy(in the form of kinetic energy in this case) then you will also gain mass.

[w=f[z]]

I might also add an admonition to not confuse mass with matter - an easy mistake to make.

 

Cheers

[foofighter]

wow nice - thanks a lot guys : )

[swansont]

E=mc^2

 

mass is energy and energy is mass. if you gain energy(in the form of kinetic energy in this case) then you will also gain mass.

 

With the caveat that you will never notice this increase in mass, since you are at rest in your own frame. It is purely a manifestation of measurement from another frame of reference, using a particular definition of mass (relativistic)

[lucaspa]

It is purely a manifestation of measurement from another frame of reference, using a particular definition of mass (relativistic)

 

If you are hit with a mass moving at relativistic speeds, it's not "purely a manifestation of measurement" You get the impact of the increased mass.

[swansont]

If you are hit with a mass moving at relativistic speeds, it's not "purely a manifestation of measurement" You get the impact of the increased mass.

 

But that's the mass of the other particle, not you. And your truncated quote changes the context of the comment. I did not say that the increased mass was purely a manifestation of measurement (as if it was an illusion), I said it was a manifestation of measurement from another frame of reference, i.e. you see it because you are in a different frame.

[lucaspa]

I said it was a manifestation of measurement from another frame of reference, i.e. you see it because you are in a different frame.

 

I understand what you are trying to say: if you were aboard a spaceship traveling at relativistic speeds, you would not notice any increased mass.

 

But this isn't just a "manifestation of measurement from another frame of reference". It's real even if the people on the spacecraft don't detect it. That increased mass produces real effects if that spacecraft collides with a planet or other object! And the effect on the spacecraft is the release of all the energy of that increased mass it had while traveling at relativistic speeds, not the "rest" mass detected by the people on board the spacecraft.

[swansont]

I understand what you are trying to say: if you were aboard a spaceship traveling at relativistic speeds, you would not notice any increased mass.

 

But this isn't just a "manifestation of measurement from another frame of reference". It's real even if the people on the spacecraft don't detect it. That increased mass produces real effects if that spacecraft collides with a planet or other object! And the effect on the spacecraft is the release of all the energy of that increased mass it had while traveling at relativistic speeds, not the "rest" mass detected by the people on board the spacecraft.

 

But to the occupants, the spacecraft isn't travelling at relativistic speeds, the object with which it collides is. The collision, from their point of view, will make perfect physics sense using their rest mass.

[Sayonara]

It will still hurt though.

[swansont]

That's relative, too. It won't hurt me at all if you're the one who crashes.

[lucaspa]

But to the occupants, the spacecraft isn't travelling at relativistic speeds, the object with which it collides is. The collision, from their point of view, will make perfect physics sense using their rest mass.

 

I have to think about that one. The energy released is calculated based on the relativistic mass of the spacecraft, That energy has to be the same for both frames of references. But if, as you say, the people in the spacecraft think the planet is travelling at relativistic velocity, then the collision should release a lot more energy, shouldn't it? Because the planet would be REALLY massive. That means 2 different amounts of energy released -- one for each frame of reference. That doesn't sound right.

 

Swansout, it will hurt you if you are under the point of impact!

[timo]

I'm not sure how you want to calculate the "energy released", not even what it's supposed to be (and admittedly I don't want to think about it right now). But energy is not a conserved quantity under changes of coordinate systems - you can easily see that when you consider a particle in its rest-frame where it has no kinetic energy and a frame where it moves (where it then has kinetic energy). There's a combined structure of energy and momentum called "4-momentum" that is conserved, but in a slightly different manner: It's a vector with 4 entries. The entries themselves change but the vector itself is considered unchanged - imagine describing an arrow on the wall with two different cordinate systems like {"up", "left"} and {"down", "right"}: The entries of your descriptions are different (the respecitve negative of each other) but the arrow is the same in both cases.

 

btw.: It seems you used this coordinate transformation on the 2nd letter "n" in Swansont's name

[abskebabs]

btw.: It seems you used this coordinate transformation on the 2nd letter "n" in Swansont's name

lol, well spotted;) . Regarding 4-vectors, can we always consider them as "arrow" vectors with regard to general coordinate transformations, or do we have to start taking into account whether they are contravariant or covariant, or whether they are densities, capacities, or "ordinary"?

 

I liked the explanation using 4-vectors, and so I have a further question. I know we can work out a scalar called the "invariant interval" when taking the relativistic product of 1 spacetime displacement 4-vector with itself(or even with another, but never mind that). Could something similiar be done with 4-momentum vectors, to show a similar invariance, and hence demonstrate a relativistic principle of conversation this way?

[swansont]

That means 2 different amounts of energy released -- one for each frame of reference. That doesn't sound right.

 

But it is, as Atheist points out. I would have tried to stay away from using "conserved" when discussing the transformation, because it might cause confusion. Energy is conserved within each frame, but does not transform between frames.

 

 

And there are worse transformations for my name (transposing the 6th and 7th letters is one, and fairly common. Hmmm. Cygnus mucosi)

[abskebabs]

lol, well spotted;) . Regarding 4-vectors, can we always consider them as "arrow" vectors with regard to general coordinate transformations, or do we have to start taking into account whether they are contravariant or covariant, or whether they are densities, capacities, or "ordinary"?

 

I liked the explanation using 4-vectors, and so I have a further question. I know we can work out a scalar called the "invariant interval" when taking the relativistic product of 1 spacetime displacement 4-vector with itself(or even with another, but never mind that). Could something similiar be done with 4-momentum vectors, to show a similar invariance, and hence demonstrate a relativistic principle of conversation this way?

Could someone please answer the questions I have given?

[timo]

Abskebabs, the problem with your questions is that a serious answer would probably be too long for a forum post (and that my article explaining at least the very start of it was lost with WiSci). So a very compressed reply:

Regarding 4-vectors, can we always consider them as "arrow" vectors with regard to general coordinate transformations ...

Kind of "yes". GR bases on "differential geometry", hence thinking of vectors as arrows (tangential vectors on spacetime, actually) imho greatly helps in understanding them.

, or do we have to start taking into account whether they are contravariant

A "contravariant vector" strictly speaking is not a vector but its coordinates in a given base (which is usually not mentioned explcitely). The name comes from that for the vector to remain the same under a base transformation, the coordinates must transform inversely to the base. Note: If the coordinate base is given or obvious, then of course the four numbers of the "covariant vector" are a sufficient description for a vector.

or covariant, ...

Pretty much the same as with contravariant vectors except that they are the coordinates in the reciprocal base (which is pretty much the same as the reciprocal base in solid state physics except for the factor 2pi and a different metric).

 

I know we can work out a scalar called the "invariant interval" when taking the relativistic product of 1 spacetime displacement 4-vector with itself(or even with another, but never mind that). Could something similiar be done with 4-momentum vectors, to show a similar invariance, and hence demonstrate a relativistic principle of conversation this way?

I don't exactly get the question, but the pseudo-scalar product (which is often just called "scalar product") of two vectors is an invariant scalar, regardless of being the scalar product of the vector with itself or with some other vector. That's similar to the angle between two lines drawn on a piece of paper being independent of the coordinate you use to describe points on this piece of paper (in fact it's even independent on whether you use a coordinate system at all or not - you just can't calculate anything without a coordinate system).

 

The result of the product of a momentum 4-vector with itself is the rest-mass squared:

[math] \left< p | p \right>_{\text{pseudo}} = \underbrace{E^2 - {|\vec p|}^2}_{\text{requires a coordinate-system to make sense}} = \underbrace{{m_0}^2 }_{\text{always true, no cs needed}}[/math]

[abskebabs]

I don't exactly get the question, but the pseudo-scalar product (which is often just called "scalar product") of two vectors is an invariant scalar, regardless of being the scalar product of the vector with itself or with some other vector. That's similar to the angle between two lines drawn on a piece of paper being independent of the coordinate you use to describe points on this piece of paper (in fact it's even independent on whether you use a coordinate system at all or not - you just can't calculate anything without a coordinate system).

 

The result of the product of a momentum 4-vector with itself is the rest-mass squared:

[math] \left< p | p \right>_{\text{pseudo}} = \underbrace{E^2 - {|\vec p|}^2}_{\text{requires a coordinate-system to make sense}} = \underbrace{{m_0}^2 }_{\text{always true, no cs needed}}[/math]

 

What I meant was; As the relativistic "scalar product" of a 4 vector with itself is an invariant scalar(as shown by the spacetime invariant interval), thereby the same thing should be true with a 4-momentum vector. Accoringly, if I take the same 4 vector measured from another frame of reference, therby containing different space and time components; I will get the same invariant scalar when I take its "scalar product" with itself. I can see now, along with what you have put up in your last post, that this has to be true, I guess it's actually trivial!

[lucaspa]

But energy is not a conserved quantity under changes of coordinate systems - you can easily see that when you consider a particle in its rest-frame where it has no kinetic energy and a frame where it moves (where it then has kinetic energy). There's a combined structure of energy and momentum called "4-momentum" that is conserved, but in a slightly different manner: It's a vector with 4 entries. The entries themselves change but the vector itself is considered unchanged - imagine describing an arrow on the wall with two different cordinate systems like {"up", "left"} and {"down", "right"}: The entries of your descriptions are different (the respecitve negative of each other) but the arrow is the same in both cases.

 

Atheist, Swansont, I'm having a difficult time seeing this. Can you guys please walk me thru an example calculation:

 

A 1 gm sphere A moving at 0.9c relative to a 100 gm sphere B.

 

So, can you do the energy calculation based on the perspective of A where A is at "rest" and B is moving at 0.9c?

 

Then the calculation from the perspective of B where B is at rest and A is moving at 0.9c?

 

That way I can see that the energy released comes out the same.

 

Thanks

[timo]

Atheist, Swansont, I'm having a difficult time seeing this. Can you guys please walk me thru an example calculation:

A 1 gm sphere A moving at 0.9c relative to a 100 gm sphere B.

 

So, can you do the energy calculation based on the perspective of A where A is at "rest" and B is moving at 0.9c?

[math] E_A = m_A c^2 + \gamma m_B c^2, \ \gamma = \sqrt{1/0.19} [/math]

 

 

Then the calculation from the perspective of B where B is at rest and A is moving at 0.9c?

[math] E_B = \gamma m_A c^2 + m_B c^2, \ \gamma = \sqrt{1/0.19} [/math]

 

 

That way I can see that the energy released comes out the same.

I still don't see any energy released, but the total energy (before any potential crash) is different, already.

[swansont]

That way I can see that the energy released comes out the same.

 

It's not going to be the same. Total energy isn't invariant under the transformation.

[lucaspa]

[math] E_A = m_A c^2 + \gamma m_B c^2, \ \gamma = \sqrt{1/0.19} [/math]

 

Where does the 0.9c come in? Where did you get 1/0.19?

 

[math] E_B = \gamma m_A c^2 + m_B c^2, \ \gamma = \sqrt{1/0.19} [/math]

 

 

I still don't see any energy released, but the total energy (before any potential crash) is different, already.

 

Waht you have is the total energy of the system A + B, right? And that energy is different based on the perspective. That shouldn't be. Somewhere the equations are off.

 

It's not going to be the same. Total energy isn't invariant under the transformation.

 

Wait a minute! We have 1 event but in 2 different reference frames. By conservation of energy the energy released by this event is a single value. It can't be one value in the spaceship and a different value on the planet! One event.

[timo]

If you're replying to "energy isn't conserved under changes of coordinate systems" with "show me the math showing this" and then reply to the numbers with "that cannot be true because energy is conserved under changes of coordinate systems" we're stuck in a loop, I think.

 

Yes, the two values [math]E_A[/math] and [math]E_B[/math] are the values for total energy in the respective systems (you described). Yes, they are unequal. The \sqrt{1/0.19} comes from gamma = \sqrt{1/(1-v²/c²)} = \sqrt{1/(1 - 0.9²c²/c²)} = \sqrt{1/0.19} and is the Gamov-factor.

 

See it this way: Energy is conserved for any physical processes. Changes of coordinate systems are not physical processes, but merely switching to a different description.

[swansont]

[math] E_A = m_A c^2 + \gamma m_B c^2' date=' \ \gamma = \sqrt{1/0.19} [/math']

 

Where does the 0.9c come in? Where did you get 1/0.19?

 

.9^2 = 0.81

 

1-.81 = 0.19

 

 

Wait a minute! We have 1 event but in 2 different reference frames. By conservation of energy the energy released by this event is a single value. It can't be one value in the spaceship and a different value on the planet! One event.

 

Energy is conserved within a reference frame, but not in the transformation to another reference frame. What you are quoting is not conservation of energy. You are claiming that total energy is invariant under a coordinate transformation, and it isn't. (values that are, though, like rest mass, are really useful)

 

I move at 3 m/s relative to everyone. Everyone else sees me with, say, 1000J of KE. I see everyone else with a speed of 3 m/s. That's a huge amount of energy. It's not invariant.