• Support PF! Buy your school textbooks, materials and every day products Here!

Moving and not moving people using Lorentz transformation

  • #1
575
47

Homework Statement



Ok I have a moving person (primed) going 50 m/s in the positive x direction, and I have someone stationary (unprimed) observing them.
At t = 0, the moving person is at x(0) = 100m
Write an equation for the object’s position as a function of time x(t)
seen by the observer in the unprimed (stationary) reference frame.

And it says the only unknown in my expression should be the variable t.

So to be clear, the stationary reference frame is unprimed. The moving reference frame is primed.

Homework Equations



[tex]x'=\frac{x-vt}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]
[tex]x=\frac{x'-vt'}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

The Attempt at a Solution


First of all, the wording of this confuses me. From what I've been reading in the book, the first equation is for the stationary (unprimed) observer. And the primed (moving) reference frame is the second equation.
But the question asks me to write an equation for x(t) for the unprimed observer. Shouldn't that be x'(t)?

Also, since the primed reference frame is moving with the person who is moving in the primed reference frame, does that mean their position is always 0? From their perspective, it's the unprimed observer who we consider stationary who is moving.

So I get...
[tex]0=\frac{100-50t}{\sqrt{1-\frac{50^2}{(3x10^8)^2}}}[/tex]

So I replaced all the variables I knew, which was x, given in the original problem as 100. And v, given as 50. And the assumed x'(t) of 0. Is that right? I think I'm way off and not even close to understanding what I'm being asked to do here.

Thanks
 

Answers and Replies

  • #2
rude man
Homework Helper
Insights Author
Gold Member
7,706
753

Homework Statement



Homework Equations



[tex]x'=\frac{x-vt}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]
[tex]x=\frac{x'-vt'}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]
The last equation is incorrect. Let v << c and that should be obvious. Since x' = 0 then the person who is stationary in the primed system is moving in which direction as seen from the person in the unprimed system? Remember, v is the speed of the primed system as seen from the unprimed system so if v > 0 then the primed system is moving in which direction as seen by the unprimed system?

The Attempt at a Solution


First of all, the wording of this confuses me. From what I've been reading in the book, the first equation is for the stationary (unprimed) observer. And the primed (moving) reference frame is the second equation.
But the question asks me to write an equation for x(t) for the unprimed observer. Shouldn't that be x'(t)?
No. That's why it's x(t) instead of x'(t). x(t) is measured in the unprimed ("stationary"} system. x'(t)= 0 always.
Also, since the primed reference frame is moving with the person who is moving in the primed reference frame, does that mean their position is always 0?
Yes. The person is stationary in the primed system. In other words, that person is the primed system. There is no motion in the primed system so x' = 0 always.

Fix you equation for x(t) and realize v is a positive number. v is defined as the primed system moving wrt the unprimed system.

Finally, you need to write t' as a function of t (think time dilation). You want x(t), not x(t'). At t = 0 the primed person is 100m from the unprimed observer, at which time t' = 0 also. Then as t increases, how does t behave compared to t'?
 
  • #3
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,623
1,257
A slight correction: it's actually x'(t') = constant. The person moving in the unprimed frame is stationary in the primed frame, but not necessarily at the origin.

The equations you wrote down for the Lorentz transformations work for two frames where the origin of the primed and unprimed frames coincide. Following this convention, you should be able to see that x'(t') isn't equal to 0.
 
  • #4
rude man
Homework Helper
Insights Author
Gold Member
7,706
753
A slight correction: it's actually x'(t') = constant. The person moving in the unprimed frame is stationary in the primed frame, but not necessarily at the origin.

The equations you wrote down for the Lorentz transformations work for two frames where the origin of the primed and unprimed frames coincide. Following this convention, you should be able to see that x'(t') isn't equal to 0.
Big 10-4.
 
  • #5
575
47
Thanks for the response.
The last equation is incorrect. Let v << c and that should be obvious. Since x' = 0 then the person who is stationary in the primed system is moving in which direction as seen from the person in the unprimed system? Remember, v is the speed of the primed system as seen from the unprimed system so if v > 0 then the primed system is moving in which direction as seen by the unprimed system?
You're right, it's supposed to be [itex]x=\frac{x'+vt'}{\sqrt{1-\frac{v^2}{c^2}}}[/itex]
No. That's why it's x(t) instead of x'(t). x(t) is measured in the unprimed ("stationary"} system. x'(t)= 0 always.
In my book it has two formulas. One of them is [itex]x'=γ(x-vt)[/itex]. [itex]γ=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/itex]. The other is [itex]x=γ'(x'+vt')[/itex]. For the first one, it says "We require a linear transformation so that each event in system K corresponds to one, and only one, event in system K'." And that's the equation they give. The second equation they give says "We can use similar arguments from the standpoint of an observer stationed in system K' (moving reference frame) to obtain an equation similar to the previous one." So according to this book, [itex]x=γ'(x'+vt')[/itex] is from the standpoint of the observer in the moving reference frame.
So as far as the question on my homework goes, it says "equation for the objects position as a function of time x(t) seen by the observer in the unprimed reference frame." So that corresponds to the first equation of the two I just typed in this paragraph, correct?
Finally, you need to write t' as a function of t (think time dilation). You want x(t), not x(t'). At t = 0 the primed person is 100m from the unprimed observer, at which time t' = 0 also. Then as t increases, how does t behave compared to t'?
This is a bit confusing to me. So t' changes as t changes, but it changes slower. Not sure how I'd write that out in equation form. Do I use the same formulas, but solve for t'?
The person moving in the unprimed frame is stationary in the primed frame, but not necessarily at the origin.
The person moving in the unprimed frame should be stationary in the unprimed frame, since he's moving with the frame, right?
The equations you wrote down for the Lorentz transformations work for two frames where the origin of the primed and unprimed frames coincide. Following this convention, you should be able to see that x'(t') isn't equal to 0.
Wouldn't x'(t') equal zero since no matter how much time elapses in the primed frame, the person never moves, because the primed frame is moving with him?

Thanks for the responses guys.
 
  • #6
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
14,759
6,271
Either I don't understand this question in the original posting or it's a trick question.

If I understand the somwhat unclearly formulated problem correctly, it's already everything said, how the primed person is moving wrt. to the unprimed. There is nothing to calculate, just write down the formula for uniform motion ;-).
 
  • #7
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,623
1,257
Wouldn't x'(t') equal zero since no matter how much time elapses in the primed frame, the person never moves, because the primed frame is moving with him?
No, stationary doesn't imply x'=0. A person sitting at x'=0 for all t' is indeed stationary in the primed frame, but just because someone's stationary doesn't mean the person is located at the origin.
 
  • #8
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,623
1,257
Either I don't understand this question in the original posting or it's a trick question.

If I understand the somwhat unclearly formulated problem correctly, it's already everything said, how the primed person is moving wrt. to the unprimed. There is nothing to calculate, just write down the formula for uniform motion ;-).
It's kind of trick question in a sense. The idea is to get students to recognize that there's really no relativity needed to answer the question.
 
  • #9
rude man
Homework Helper
Insights Author
Gold Member
7,706
753
No, stationary doesn't imply x'=0. A person sitting at x'=0 for all t' is indeed stationary in the primed frame, but just because someone's stationary doesn't mean the person is located at the origin.
Isn't he located at the origin because he is the origin? We don't have motion within the primed system.

x = (x' + vt' )γ + 100
but x' = 0
so x = vt'γ + 100
but t = t'γ
therefore x = vt + 100.
 
Last edited:
  • #10
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,623
1,257
You're changing the equations. You can't have both ##x = \gamma(x'+vt')## and ##x = \gamma(x'+vt') + 100## be true at the same time.
 
  • #11
rude man
Homework Helper
Insights Author
Gold Member
7,706
753
You're changing the equations. You can't have both ##x = \gamma(x'+vt')## and ##x = \gamma(x'+vt') + 100## be true at the same time.
I previously acknowledged that I omitted the 100m.
 
  • #12
575
47
Ok this makes no sense to me.
Here's the equation:
[tex]x=\frac{x'+vt'}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

I'm supposed to write out an expression for x(t)... but there's no t in that equation. It's t'.
Am I supposed to be creating this expression:
[tex]x(t)=\frac{x'+vt'}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

Can I legally just turn that x into x(t)? Even though there's no t in the expression?
 
  • #13
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,623
1,257
Reread posts 6 and 8.
 
  • #14
575
47
Reread posts 6 and 8.
Oh yeah, duh, t and t' are equal.

If they weren't equal, creating such an equation would not be legal?
 
  • #15
rude man
Homework Helper
Insights Author
Gold Member
7,706
753
Ok this makes no sense to me.
Here's the equation:
[tex]x=\frac{x'+vt'}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

I'm supposed to write out an expression for x(t)... but there's no t in that equation. It's t'.
Am I supposed to be creating this expression:
[tex]x(t)=\frac{x'+vt'}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

Can I legally just turn that x into x(t)? Even though there's no t in the expression?
Yes, but you need to add 100 m to x(t) since the fellow at x' = 0 starts at 100 m from the fellow at x = 0.

You obviously must replace t' with t though. What is the relationship between t' and t? Remember I suggested 'time dilation'?
 
  • #16
rude man
Homework Helper
Insights Author
Gold Member
7,706
753
  • #17
575
47
Yes, but you need to add 100 m to x(t) since the fellow at x' = 0 starts at 100 m from the fellow at x = 0.
I was trying to figure out where to put that 100 m. I don't replace x(t) with it? I just add it to the equation? The problem says I should have only 1 variable in the expression, and that's t. So if x(t) isn't 100 m, what is it? Does the x(t) count as a variable that should be represented by a constant?
Yes, but you need to add 100 m to x(t) since the fellow at x' = 0 starts at 100 m from the fellow at x = 0.

You obviously must replace t' with t though. What is the relationship between t' and t? Remember I suggested 'time dilation'?
The relationship between t' and t is:
[tex]t'=\frac{t-\frac{v}{c^2}x}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]
And
[tex]t=\frac{t'-\frac{v}{c^2}x'}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]
So when v << c, t = t'
 
Last edited:
  • #18
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,623
1,257
You've missed the point again. Look, the problem says at t=0, the person is at x(0)=100 m. The person is moving at 50 m/s. Write down x(t). The stuff about the second frame has absolutely nothing to do with this problem.
 
  • #19
rude man
Homework Helper
Insights Author
Gold Member
7,706
753
The relationship between t' and t is:
[tex]t'=\frac{t-\frac{v}{c^2}x}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]
And
[tex]t=\frac{t'-\frac{v}{c^2}x'}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]
So when v << c, t = t'
The second equation is appropriate but incorrect, so fix it. (Actually, here it doesn't matter, but fix it anyway).

x' = 0, remember? The guy in the frame x' IS the x' frame, and he is always at x' = 0.
So, use that info to get t(t') and from there x(t). Don't forget the 100 m.
 
  • #20
575
47
You've missed the point again. Look, the problem says at t=0, the person is at x(0)=100 m. The person is moving at 50 m/s. Write down x(t). The stuff about the second frame has absolutely nothing to do with this problem.
I have to write down x(t) and replace all the variables except t with values. If x(t) is a variable, I don't know how to replace that with a value while keeping t as a variable.
The second equation is appropriate but incorrect, so fix it. (Actually, here it doesn't matter, but fix it anyway).
I keep forgetting about the +. It's correct in my notes.
So, use that info to get t(t') and from there x(t).
Is t(t') just the equation I typed, but replacing t with t(t')? I think a big part of my problem is creating these variable as a function of some other variable expressions is something I've never done in any class I've ever taken, so I'm not sure on the rules.
 
Last edited:
  • #21
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,623
1,257
Forget about relativity. It has nothing to do with this problem because both x and t are coordinates in the non-moving frame. You're not transforming between frames here.

Just use the equations you learned in plain old Newtonian kinematics.
 
  • #22
rude man
Homework Helper
Insights Author
Gold Member
7,706
753
I have to write down x(t) and replace all the variables except t with values. If x(t) is a variable, I don't know how to replace that with a value while keeping t as a variable.

I keep forgetting about the +. It's correct in my notes.

Is t(t') just the equation I typed, but replacing t with t(t')? I think a big part of my problem is creating these variable as a function of some other variable expressions is something I've never done in any class I've ever taken, so I'm not sure on the rules.
I agree that on careful reading of the problem, which I didn't do, relativity plays no part. But, I think it still serves a purpose to go thru the relativistic exercise, if for no other reason than to show complementarity with Newtonian physics.

So in that vein, I dont see why you can't seem to use your expression for t to get a relationship between t and t'. It's really very simple. Just set x' = 0.
 
Last edited:

Related Threads on Moving and not moving people using Lorentz transformation

Replies
1
Views
1K
Replies
3
Views
327
  • Last Post
Replies
8
Views
2K
Replies
1
Views
962
Replies
14
Views
2K
Replies
1
Views
1K
Replies
0
Views
1K
Replies
1
Views
1K
Replies
1
Views
491
Replies
2
Views
2K
Top