Magnitude of a Four Velocity

  • Thread starter Hertz
  • Start date
  • #1
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8

Homework Statement



Prove that the magnitude of the general four velocity U = (γc, γvx, γvy, γvz) is always equal to c.

The Attempt at a Solution


|U| =

=[itex]\sqrt{(\gamma c)^{2}+(\gamma v_{x})^{2}+(\gamma v_{y})^{2}+(\gamma v_{z})^{2}}[/itex]
=[itex]\gamma\sqrt{c^{2}+v_{x}^{2}+v_{y}^{2}+v_{z}^{2}}[/itex]
=[itex]\gamma\sqrt{c^{2}+v^{2}}[/itex]
=[itex]c\gamma\sqrt{1+\frac{v^{2}}{c^{2}}}[/itex]
=[itex]c\frac{\sqrt{1+\frac{v^{2}}{c^{2}}}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}[/itex]

As you can see, somehow I'm off by a minus :\ Can someone help me out here? I'm guessing these two square roots are supposed to cancel somehow.

Any help is appreciated :\ Been staring at this for a while now

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Answers and Replies

  • #2
20,875
4,547
In your first equation, there should be a minus sign in front of the ct term. In evaluating the square of the magnitude of a 4 vector, the square of the time component gets a minus sign (as in the Minkowski metric).
 
  • #3
180
8
In your first equation, there should be a minus sign in front of the ct term. In evaluating the square of the magnitude of a 4 vector, the square of the time component gets a minus sign (as in the Minkowski metric).
Okie dokie..

It seems rather arbitrary in my opinion.. What should I write next to my work in order to rationalize this step? Why is finding the magnitude of the four velocity any different than finding the magnitude of any other four vector?

-edit
Also, do I have the general four velocity written wrong? I'm looking at wikipedia right now and it says the first component should be ct = (gamma)c(tau).. Why does my teacher have it written here as (gamma)c?

Sorry, I know I should be more competent with four velocity at this point but this is the first problem I've encountered dealing with it and I swear info for this stuff is impossible to find.
 
Last edited:
  • #4
20,875
4,547
Okie dokie..

It seems rather arbitrary in my opinion.. What should I write next to my work in order to rationalize this step? Why is finding the magnitude of the four velocity any different than finding the magnitude of any other four vector?

-edit
Also, do I have the general four velocity written wrong? I'm looking at wikipedia right now and it says the first component should be ct = (gamma)c(tau).. Why does my teacher have it written here as (gamma)c?

Sorry, I know I should be more competent with four velocity at this point but this is the first problem I've encountered dealing with it and I swear info for this stuff is impossible to find.
The relationship you wrote for the 4 velocity applies to Euclidean space. Relativistic 4D space-time is not Euclidean. The Lorentz Transformation tells us that the metric for space-time has a minus sign in front of the square of the time component. This is a fundamental geometric feature of 4D space-time. If you use the Euclidean relationship for space time, you will get the wrong answer.

Regarding your second question, ct is not velocity, it is distance. The article in Wiki must be referring to distances rather than velocities.
 
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  • #5
180
8
Ok thank you. I have looked into it a bit more and I now realize that the time coordinate is imaginary. It's still pretty abstract in my opinion, I really wish I had more time to research it and get a better understanding of why this is the case, but I need to get this homework done. I'll look into it in more detail in future physics classes I'm sure.

Finally, why is it ok to make all three of the other coordinates negative rather than time? I realize that when you square the components you can either end up with (-, +, +, +) or (+, -, -, -) but to go in between these wouldn't you have to add a negative 1 under the square root, and consequently, an i on the outside of the square root? If I did this, I would end up with the magnitude as being ci.
 
  • #6
UltrafastPED
Science Advisor
Gold Member
1,912
216
The spacetime metric derives from this truth: for a light pulse taken as an event in spacetime all observers will agree with the observation (in their own inertial reference frame, and using their own "true" clocks and rulers) that:

(ct)^2 = x^2 + y^2 + z^2.

When expressed as a 4-vector you have (ct, x, y, z), and the length of this four vector is
s^2 = -(ct)^2 + x^2 + y^2 + z^2 This is the (-+++) metric; (+---) also works, but you must be consistent.

Minkowski decided to use the Euclidean metric, and thus stuck i=sqrt(-1) in front of the ct. But this simplification doesn't work with General Relativity, so you may as well get used to the notion of a metric - it goes with the geometry of spacetime.

The invariant "spacetime interval" was calculated from s^2 above; for light it is always zero; but it can also be positive or negative ... and the interpretation of these values depends upon the convention followed. But one is always "inside the light cone", and the other is always outside. The "null geodesics", the lines where s^2=0, are the boundaries of the light cone.

When you have time read Taylor & Wheeler's "Spacetime Physics" - over winter break or even better, in the summer - then you can work the problems.
 
  • #7
20,875
4,547
Ok thank you. I have looked into it a bit more and I now realize that the time coordinate is imaginary. It's still pretty abstract in my opinion, I really wish I had more time to research it and get a better understanding of why this is the case, but I need to get this homework done. I'll look into it in more detail in future physics classes I'm sure.

Finally, why is it ok to make all three of the other coordinates negative rather than time? I realize that when you square the components you can either end up with (-, +, +, +) or (+, -, -, -) but to go in between these wouldn't you have to add a negative 1 under the square root, and consequently, an i on the outside of the square root? If I did this, I would end up with the magnitude as being ci.
UltrafastPED did a nice job of explaining things. I would just like to add a couple of extra comments. Resorting to imaginary coordinates is a practice that is frowned upon modern day and, as UltrafastPED pointed out, doesn't work with general relativity. Secondly, the two opposite "sign signatures" you referred to are both used in relativity texts and literature. Both are acceptable, and lead to identical results when applied to practical problems.
 
  • #8
jpbbrito

Homework Statement



Prove that the magnitude of the general four velocity U = (γc, γvx, γvy, γvz) is always equal to c.

The Attempt at a Solution


|U| =

=[itex]\sqrt{(\gamma c)^{2}+(\gamma v_{x})^{2}+(\gamma v_{y})^{2}+(\gamma v_{z})^{2}}[/itex]
=[itex]\gamma\sqrt{c^{2}+v_{x}^{2}+v_{y}^{2}+v_{z}^{2}}[/itex]
=[itex]\gamma\sqrt{c^{2}+v^{2}}[/itex]
=[itex]c\gamma\sqrt{1+\frac{v^{2}}{c^{2}}}[/itex]
=[itex]c\frac{\sqrt{1+\frac{v^{2}}{c^{2}}}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}[/itex]

As you can see, somehow I'm off by a minus :\ Can someone help me out here? I'm guessing these two square roots are supposed to cancel somehow.

Any help is appreciated :\ Been staring at this for a while now

Homework Statement





Homework Equations





The Attempt at a Solution


1. Homework Statement [/B]



Homework Equations





The Attempt at a Solution


1. Homework Statement [/B]



Homework Equations





The Attempt at a Solution


1. Homework Statement [/B]



Homework Equations





The Attempt at a Solution


1. Homework Statement [/B]



Homework Equations





The Attempt at a Solution


1. Homework Statement [/B]



Homework Equations





The Attempt at a Solution


1. Homework Statement [/B]



Homework Equations





The Attempt at a Solution


1. Homework Statement [/B]



Homework Equations





The Attempt at a Solution


1. Homework Statement [/B]



Homework Equations





The Attempt at a Solution

Use the metric to calculate the U^{2} = \eta_{\mu \nu} U^{\mu} U^{\nu}.
 
  • #9
jpbbrito
Use the metric to calculate the U^{2} = \eta_{\mu \nu} U^{\mu} U^{\nu}
 
  • #10
20,875
4,547
Use the metric to calculate the U^{2} = \eta_{\mu \nu} U^{\mu} U^{\nu}
There have been no responses to this thread in practically 5 years. The OP was last seen almost 9 months ago. This thread is hereby closed.
 

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