- #31
Virous
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I just copied the table as it is. It is not mine. And I don`t see any mistakes in it.
Please explain what is wrong with my relativistic momentum problem
- Thread starter Virous
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- #32
xox
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Virous said:
Before: \gamma(v_A) \vec{v_A}+\gamma(v_B) \vec{v_B}
After: \gamma(v'_A) \vec{v'_A}+\gamma(v'_B) \vec{v'_B}
You have everything correct, except the way you calculate \gamma(v_B), \gamma(v'_B).
I explained the error.
- #33
Virous
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No, I mean, please, expand it (gammas). And it seems your forgot masses.
- #34
xox
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PAllen said:
He's projecting
\gamma(v_A) \vec{v_A}+\gamma(v_B) \vec{v_B}=\gamma(v'_A) \vec{v'_A}+\gamma(v'_B) \vec{v'_B}
on the x and y axis, so his using of the components is correct. His error is in calculating the \gamma(v_B),\gamma(v'_B)
- #35
xox
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The masses cancel out because both particles have the same rest mass.Virous said:
- #36
Virous
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p=v_{b}m\gamma (v_{b})=\frac{v_{b}m}{\sqrt{1-(\frac{v_{b}^2}{c^2})}}
This is my equation for initial momentum of the ball 2. (The one I used). With this equation the momentum is not conserved. Where exactly is the mistake?
This is my equation for initial momentum of the ball 2. (The one I used). With this equation the momentum is not conserved. Where exactly is the mistake?
True, but you can`t cancel them out before equating total momentum before and after.xox said:
- #37
PeterDonis
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Virous, rather than try to unpack your notation, which I (like others) am having trouble understanding, perhaps it would help to show how I would work this problem, using standard 4-vector notation. Each ball is described by a 4-momentum vector, which is just its rest mass ##m## times its 4-velocity ##u^a## (where the index ##a## denotes the coordinate components of the 4-vector, i.e., the components of ##u^a## are ##u^t##, ##u^x##, ##u^y##, and ##u^z##). The ##t## component of 4-momentum in a given frame gives the object's energy in that frame, and the ##x##, ##y##, and ##z## components give the components of its momentum. So checking conservation of momentum before and after just amounts to checking the appropriate components of the 4-momentum before and after. Since the invariant mass ##m## of each ball is the same throughout, we can divide through by it and just work with the 4-velocity vectors.
I won't give a lot of explanation of what follows; the formalism I'm using is described in most relativity textbooks. I'll just work through it quickly so you can see how it goes.
In the unprimed frame, the two 4-velocity vectors before the collision are (giving the ##t##, ##x##, and ##y## components; all ##z## components are zero):
$$
u_1^a = \left( \frac{1}{\sqrt{1 - a^2 - b^2}}, \frac{a}{\sqrt{1 - a^2 - b^2}}, \frac{b}{\sqrt{1 - a^2 - b^2}} \right)
$$
$$
u_2^a = \left( \frac{1}{\sqrt{1 - a^2 - b^2}}, \frac{- a}{\sqrt{1 - a^2 - b^2}}, \frac{- b}{\sqrt{1 - a^2 - b^2}} \right)
$$
After the collision, the two 4-velocity vectors are:
$$
u_1^a = \left( \frac{1}{\sqrt{1 - a^2 - b^2}}, \frac{a}{\sqrt{1 - a^2 - b^2}}, \frac{- b}{\sqrt{1 - a^2 - b^2}} \right)
$$
$$
u_2^a = \left( \frac{1}{\sqrt{1 - a^2 - b^2}}, \frac{- a}{\sqrt{1 - a^2 - b^2}}, \frac{b}{\sqrt{1 - a^2 - b^2}} \right)
$$
Obviously the ##x## and ##y## components of the two vectors cancel both before and after, so momentum is conserved. (Btw, the sum of the two ##t## components is also the same before and after, so energy is conserved as well.)
Now we want to Lorentz transform these 4-vectors to the primed frame, which is moving in the positive ##x## direction at speed ##a##. The nice thing about 4-vectors is that they all Lorentz transform the same; a LT in the ##x## direction only changes the ##t## and ##x## components, and it does so in the same way you're used to from basic relativity physics. That is, we have ##u'^t = \gamma \left( u^t - v u^x \right)##, and ##u'^x = \gamma \left( u^x - v u^t \right)##, where ##v## is the relative velocity (##a## in this case) and ##\gamma = 1 / \sqrt{1 - v^2}## (note that I am using units in which ##c = 1## for simplicity); and ##u^y## and ##u^z## are unchanged.
Now, if you look at the before and after 4-vectors given above, you will notice something: the ##t## and ##x## components are *the same* before and after in the unprimed frame. And since the LT only changes the ##t## and ##x## components (i.e., it doesn't "mix in" any others), the same thing will be true in the primed frame: the ##t'## and ##x'## components will be the same both before and after. So it's obvious that energy and ##x## momentum will be conserved in the primed frame if they are conserved in the unprimed frame (which of course they are). And since the ##y## components are unchanged by the LT, ##y## momentum will also be conserved in the primed frame.
So we really don't even need to work through the details of transforming into the primed frame; but I'll do it anyway for completeness. Applying the LT, in the primed frame we have these two 4-velocity vectors before the collision (where I have refrained from simplifying the expressions so you can see how they arise from the general LT formula above):
$$
u_1^a = \left( \frac{1 - a^2}{\sqrt{1 - a^2 - b^2} \sqrt{1 - a^2}}, \frac{a - a}{\sqrt{1 - a^2 - b^2} \sqrt{1 - a^2}}, \frac{b}{\sqrt{1 - a^2 - b^2}} \right)
$$
$$
u_2^a = \left( \frac{1 + a^2}{\sqrt{1 - a^2 - b^2} \sqrt{1 - a^2}}, \frac{- a - a}{\sqrt{1 - a^2 - b^2} \sqrt{1 - a^2}}, \frac{- b}{\sqrt{1 - a^2 - b^2}} \right)
$$
Simplifying now, we get
$$
u_1^a = \left( \frac{\sqrt{1 - a^2}}{\sqrt{1 - a^2 - b^2}}, 0, \frac{b}{\sqrt{1 - a^2 - b^2}} \right)
$$
$$
u_2^a = \left( \frac{1 + a^2}{\sqrt{1 - a^2 - b^2} \sqrt{1 - a^2}}, \frac{- 2a}{\sqrt{1 - a^2 - b^2} \sqrt{1 - a^2}}, \frac{- b}{\sqrt{1 - a^2 - b^2}} \right)
$$
After the collision, all that changes is that the ##y## components swap, as before, so we have:
$$
u_1^a = \left( \frac{\sqrt{1 - a^2}}{\sqrt{1 - a^2 - b^2}}, 0, \frac{- b}{\sqrt{1 - a^2 - b^2}} \right)
$$
$$
u_2^a = \left( \frac{1 + a^2}{\sqrt{1 - a^2 - b^2} \sqrt{1 - a^2}}, \frac{- 2a}{\sqrt{1 - a^2 - b^2} \sqrt{1 - a^2}}, \frac{b}{\sqrt{1 - a^2 - b^2}} \right)
$$
Again, it's obvious that the ##y## momentum cancels, and that the total ##x## momentum, while it is no longer zero in this frame, is the same both before and after. (And, of course, so is total energy, although it is larger in this frame than in the unprimed frame.)
I won't give a lot of explanation of what follows; the formalism I'm using is described in most relativity textbooks. I'll just work through it quickly so you can see how it goes.
In the unprimed frame, the two 4-velocity vectors before the collision are (giving the ##t##, ##x##, and ##y## components; all ##z## components are zero):
$$
u_1^a = \left( \frac{1}{\sqrt{1 - a^2 - b^2}}, \frac{a}{\sqrt{1 - a^2 - b^2}}, \frac{b}{\sqrt{1 - a^2 - b^2}} \right)
$$
$$
u_2^a = \left( \frac{1}{\sqrt{1 - a^2 - b^2}}, \frac{- a}{\sqrt{1 - a^2 - b^2}}, \frac{- b}{\sqrt{1 - a^2 - b^2}} \right)
$$
After the collision, the two 4-velocity vectors are:
$$
u_1^a = \left( \frac{1}{\sqrt{1 - a^2 - b^2}}, \frac{a}{\sqrt{1 - a^2 - b^2}}, \frac{- b}{\sqrt{1 - a^2 - b^2}} \right)
$$
$$
u_2^a = \left( \frac{1}{\sqrt{1 - a^2 - b^2}}, \frac{- a}{\sqrt{1 - a^2 - b^2}}, \frac{b}{\sqrt{1 - a^2 - b^2}} \right)
$$
Obviously the ##x## and ##y## components of the two vectors cancel both before and after, so momentum is conserved. (Btw, the sum of the two ##t## components is also the same before and after, so energy is conserved as well.)
Now we want to Lorentz transform these 4-vectors to the primed frame, which is moving in the positive ##x## direction at speed ##a##. The nice thing about 4-vectors is that they all Lorentz transform the same; a LT in the ##x## direction only changes the ##t## and ##x## components, and it does so in the same way you're used to from basic relativity physics. That is, we have ##u'^t = \gamma \left( u^t - v u^x \right)##, and ##u'^x = \gamma \left( u^x - v u^t \right)##, where ##v## is the relative velocity (##a## in this case) and ##\gamma = 1 / \sqrt{1 - v^2}## (note that I am using units in which ##c = 1## for simplicity); and ##u^y## and ##u^z## are unchanged.
Now, if you look at the before and after 4-vectors given above, you will notice something: the ##t## and ##x## components are *the same* before and after in the unprimed frame. And since the LT only changes the ##t## and ##x## components (i.e., it doesn't "mix in" any others), the same thing will be true in the primed frame: the ##t'## and ##x'## components will be the same both before and after. So it's obvious that energy and ##x## momentum will be conserved in the primed frame if they are conserved in the unprimed frame (which of course they are). And since the ##y## components are unchanged by the LT, ##y## momentum will also be conserved in the primed frame.
So we really don't even need to work through the details of transforming into the primed frame; but I'll do it anyway for completeness. Applying the LT, in the primed frame we have these two 4-velocity vectors before the collision (where I have refrained from simplifying the expressions so you can see how they arise from the general LT formula above):
$$
u_1^a = \left( \frac{1 - a^2}{\sqrt{1 - a^2 - b^2} \sqrt{1 - a^2}}, \frac{a - a}{\sqrt{1 - a^2 - b^2} \sqrt{1 - a^2}}, \frac{b}{\sqrt{1 - a^2 - b^2}} \right)
$$
$$
u_2^a = \left( \frac{1 + a^2}{\sqrt{1 - a^2 - b^2} \sqrt{1 - a^2}}, \frac{- a - a}{\sqrt{1 - a^2 - b^2} \sqrt{1 - a^2}}, \frac{- b}{\sqrt{1 - a^2 - b^2}} \right)
$$
Simplifying now, we get
$$
u_1^a = \left( \frac{\sqrt{1 - a^2}}{\sqrt{1 - a^2 - b^2}}, 0, \frac{b}{\sqrt{1 - a^2 - b^2}} \right)
$$
$$
u_2^a = \left( \frac{1 + a^2}{\sqrt{1 - a^2 - b^2} \sqrt{1 - a^2}}, \frac{- 2a}{\sqrt{1 - a^2 - b^2} \sqrt{1 - a^2}}, \frac{- b}{\sqrt{1 - a^2 - b^2}} \right)
$$
After the collision, all that changes is that the ##y## components swap, as before, so we have:
$$
u_1^a = \left( \frac{\sqrt{1 - a^2}}{\sqrt{1 - a^2 - b^2}}, 0, \frac{- b}{\sqrt{1 - a^2 - b^2}} \right)
$$
$$
u_2^a = \left( \frac{1 + a^2}{\sqrt{1 - a^2 - b^2} \sqrt{1 - a^2}}, \frac{- 2a}{\sqrt{1 - a^2 - b^2} \sqrt{1 - a^2}}, \frac{b}{\sqrt{1 - a^2 - b^2}} \right)
$$
Again, it's obvious that the ##y## momentum cancels, and that the total ##x## momentum, while it is no longer zero in this frame, is the same both before and after. (And, of course, so is total energy, although it is larger in this frame than in the unprimed frame.)
Last edited:
- #38
xox
- 192
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Virous said:
Look at your v^2_b. It has BOTH an x AND an y component when you calculate "gamma". You are using only the y component. THAT is your error.
Yes, irrelevant in finding your error.
- #39
PeterDonis
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Virous said:
In this particular case, since the rest masses of both balls are equal, you can; they drop out everywhere if the computation is done correctly. (You'll note that that's what I did in my previous post.)
- #40
Virous
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xox
I tried that as well. It doesn`t change anything, since even if you add the horizontal component there, momentum will still be not conserved!
I tried that as well. It doesn`t change anything, since even if you add the horizontal component there, momentum will still be not conserved!
- #41
Virous
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PeterDonis
Thanks for your large post and spending time on me! I`ll try it out later as well. But now the whole point is to understand where am I wrong.
Thanks for your large post and spending time on me! I`ll try it out later as well. But now the whole point is to understand where am I wrong.
- #42
PAllen
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xox said:
Ok, I'll buy that.
- #43
Virous
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xox, Probably, I am getting annoying, so you`re using capitals. I am really-really sorry for that. I`m just getting into it. Please, write the equation of momentum of the ball B (2) substituting everything.
- #44
PeterDonis
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Virous said:
Well, my computation suggests one obvious possibility: a Lorentz transform in the ##x## direction should not affect momentum in the ##y## direction. (It will affect *total* momentum because that's a vector sum of the ##x## and ##y## momentum; but when looking at components, only the ##x## momentum should change.) So I would guess that you've made an error in transforming the ##y## components to the primed frame.
- #45
Virous
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PeterDonis, vertical velocity in the moving frame depends on the horizontal velocity in the rest frame, so the momentum should change as well. Isn`t it?
- #46
PeterDonis
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Virous said:
It depends on how you look at it. If you look at my 4-vector formulas, you will note that the ##y## components in both frames are identical expressions--meaning that they would be numerically identical once specific numbers are plugged in for all the variables. So in the 4-vector formalism, confirming ##y## momentum conservation is simple.
What you are essentially trying to compute in your formulas are the *ratios* of 4-vector components; for example, your ##y## component of ordinary velocity is the ratio ##u^y / u^t## in my notation. Those ratios do change when you change frames, yes; but trying to compute things using them is just introducing extra steps that should end up cancelling each other out if the computation is done correctly.
In other words, your ordinary velocity ##y## component is ##v_y = u^y / u^t## in my notation: but your ##y## momentum is ##m \gamma(v) v_y##, where ##\gamma(v) = 1 / \sqrt{1 - v_x^2 - v_y^2}##. And if you look at my formulas, you will see that ##\gamma(v) = u^t##; so your ##y## momentum should end up being ##m u^t v_y = m u^t u^y / u^t = m u^y##, which is unchanged, as my computation shows. So basically, you're computing a change in the ##y## component of ordinary velocity in the primed frame; but then you should be computing an exactly compensating change in how the ##y## components of momentum and ordinary velocity are related in the primed frame. So I think your mistake lies somewhere in that computation; two things that should be exactly cancelling each other out are not, because you've made an error in there somewhere.
The above is why the 4-vector formalism is so nice; you don't have to worry about trying to figure out how all the ratios change from frame to frame, the formalism keeps track of it all for you.
- #47
PAllen
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Using the formula I gave for non-colinear velocities, and given that the lower ball is has velocity:
(a,b)
and S' has velocity (a,0), all given in S, then the gamma for this ball in S' (that is (a,b) - (a,0) relativistically, non-colinear) is:
√(1-a2)/√(1-a2-b2)
This is just algebra (and elemantary trig) from my formula. Note, it is in agreement with Peter's post.
I cannot tell from your notations, whether you agree with this. Can you answer whether you agree with this as gamma for the bottom ball in S'?
(a,b)
and S' has velocity (a,0), all given in S, then the gamma for this ball in S' (that is (a,b) - (a,0) relativistically, non-colinear) is:
√(1-a2)/√(1-a2-b2)
This is just algebra (and elemantary trig) from my formula. Note, it is in agreement with Peter's post.
I cannot tell from your notations, whether you agree with this. Can you answer whether you agree with this as gamma for the bottom ball in S'?
- #48
Virous
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Probably I`m getting something absolutely wrong. Shouldn`t the following be the relativistic momentum formula?
p=\frac{mu}{\sqrt{1-(u/c)^2 }}
Can you now, please, just take velocity values from my table and substitute them into the equation correctly?
p=\frac{mu}{\sqrt{1-(u/c)^2 }}
Can you now, please, just take velocity values from my table and substitute them into the equation correctly?
- #49
Virous
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xox, I redid the computation as you told me - here is the result:
(Total momentum before on the left, total momentum after on the right)
u1, u2 - vertical components of balls` velocities before.
u3,u4 - vertical comoponents after
Sorry for not using special tegs. What I did is I added to u2 and u4 their horizontal components as they are given in the table.
Total before is still not equal to total after (they have opposite signs instead). What is wrong now?
(Total momentum before on the left, total momentum after on the right)
u1, u2 - vertical components of balls` velocities before.
u3,u4 - vertical comoponents after
Sorry for not using special tegs. What I did is I added to u2 and u4 their horizontal components as they are given in the table.
Total before is still not equal to total after (they have opposite signs instead). What is wrong now?
- #50
PeterDonis
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Virous said:
This works if you're just treating ##u## as a vector. But if you're working with components, you need to carefully distinguish between the component values and the squared magnitude of ##u## itself. The clearest way to do that is to write out ##u^2## in terms of components, so we have:
$$
p_x = \frac{m u_x}{\sqrt{1 - u_x^2 - u_y^2 - u_z^2}}
$$
$$
p_y = \frac{m u_y}{\sqrt{1 - u_x^2 - u_y^2 - u_z^2}}
$$
$$
p_z = \frac{m u_z}{\sqrt{1 - u_x^2 - u_y^2 - u_z^2}}
$$
It doesn't look like you have been doing this correctly in your formulas.
Last edited:
- #51
xox
- 192
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This is definitely not what I have told you.Virous said:xox, I redid the computation as you told me - here is the result:
(Total momentum before on the left, total momentum after on the right)
u1, u2 - vertical components of balls` velocities before.
u3,u4 - vertical comoponents after
Sorry for not using special tegs. What I did is I added to u2 and u4 their horizontal components as they are given in the table.
Total before is still not equal to total after (they have opposite signs instead). What is wrong now?
In the unprimed frame, before the collision:
p_{Ax}=\frac{mu_{Ax}}{\sqrt{1-\frac{u_{Ax}^2+u_{Ay}^2}{c^2}}}
p_{Ay}=\frac{mu_{Ay}}{\sqrt{1-\frac{u_{Ax}^2+u_{Ay}^2}{c^2}}}
p_{Bx}=\frac{mu_{Bx}}{\sqrt{1-\frac{u_{Bx}^2+u_{By}^2}{c^2}}}
p_{By}=\frac{mu_{By}}{\sqrt{1-\frac{u_{Bx}^2+u_{By}^2}{c^2}}}
You calculate the above y incorrectly because you are inadvertently dropping u_{x} in the calculation of your \gamma for p_{Ay}, p_{By}.
In the unprimed frame, after the collision:
p'_{Ax}=\frac{mu'_{Ax}}{\sqrt{1-\frac{u_{Ax}^{'2}+u_{Ay}^{'2}}{c^2}}}
p'_{Ay}=\frac{mu'_{Ay}}{\sqrt{1-\frac{u_{Ax}^{'2}+u_{Ay}^{'2}}{c^2}}}
p'_{Bx}=\frac{mu'_{Bx}}{\sqrt{1-\frac{u_{Bx}^{'2}+u_{By}^{'2}}{c^2}}}
p'_{By}=\frac{mu'_{By}}{\sqrt{1-\frac{u_{Bx}^{'2}+u_{By}^{'2}}{c^2}}}
You calculate the above y components incorrectly because you are inadvertently dropping u'_{x} in the calculation of your \gamma for p'_{Ay}, p'_{By}. You need to start to be able to figure out your mistakes on your own after so many explanations are given.
If you look at the calculations that Peter Donis did for you, using 4 vectors and you ignore the first component (in "t") you will notice the correct formulas. Hopefully, you will notice how his \gamma differs from yours.
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- #52
PAllen
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Virous said:
If by 'your table', you mean image 4 in you original post, I either don't understand it or disagree with it. The understanding issue is I still don't see where you define gamma and beta as used in those formulas. Also, in terms of simple (a,b) as defined in image 3, the Y coordinate velocity of ball 1 in S' should simply be:
b/√(1-a2)
(it's X coordinate velocity is zero, by construction, consistent with your image 4).
(In case it's unclear, I am taking (a,b) from your image 3 as given in units of c=1).
Last edited:
- #53
Virous
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xox, that is exactly what I did. Your formulae:
Total momentum before:
Total after:
It is absolutely clear, that they are not equal!
Total momentum before:
Total after:
It is absolutely clear, that they are not equal!
- #54
Virous
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I`m sorry. Probably, I`m entirely stupid and can`t get a simple point. But it doesn`t bring me any closer to a solution of my problem.
- #55
Virous
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PAllen, Beta is defined as (a/c) and gamma as (1-√(1-(a/c)^2) in ordinary units :)
- #56
xox
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Virous said:xox, that is exactly what I did. Your formulae:
Total momentum before:
Total after:
It is absolutely clear, that they are not equal!
Of course they are not, you are adding the x components with the y components, this is wrong. The conservation says:
p_{Ax}+p_{Bx}=p'_{Ax}+p'_{Bx}
p_{Ay}+p_{By}=p'_{Ay}+p'_{By}
Your p_{Before},p_{After} are wrong since you are adding the x and y components.
- #57
Virous
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xox, if a=b and c=d, than a+c=b+d isn`t it?
P.S. In the second image it should be Pafter. Sorry :)
P.S. In the second image it should be Pafter. Sorry :)
- #58
xox
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Virous said:
Yes, but this doesn't mean that you should be doing this. Try doing it on a component by component basis.
- #59
Virous
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I tried many times in all the possible combinations. And as I told you before, x is conserved, but y is not. And my equations in the top show this. Because if a+b is not equal to b+d, as it is on the images, than there is no way for a to be equal to c together with b=d!
- #60
xox
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Virous said:
Well, we found that your initial gammas were incorrect. Why don't you write the new equations on a component basis and we'll see easier what other mistake you may have.
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