Worldlines and spacetime diagrams

In summary: Yes, the spacetime displacement of this emission event with respect to the origin of S is given by the four vector\textbf{s}_E=ct_E\textbf{e}_0+(c^2/a)\textbf{e}_1In summary, the observer's worldline expressed in terms of the coordinates of an inertial reference frame is given by: t(\tau)=(c/a)\sinh(a\tau/c)x(\tau)=(c^2/a)\sinh(a\tau/c). The radio beacon emitted at time tE has the spacetime cordinates (ct_E,x_E)=(ct
  • #1
alc95
7
0

Homework Statement


An observer on the bridge of a spaceship is undergoing a proper acceleration a, so that the
observer’s worldline expressed in terms of the coordinates of an inertial reference frame S
is given by
[tex]t(\tau)=(c/a)\sinh(a\tau/c)[/tex][tex]x(\tau)=(c^2/a)\sinh(a\tau/c)[/tex]
(a) Draw on a spacetime diagram for the reference frame S the worldline of the observer
for tau > 0.
At the instant tau = 0 a radio beacon is dropped off by the observer. At the instant of being
dropped off, the beacon is at rest relative to the observer. At this instant the beacon also
starts emitting at regular intervals (according to a clock on the beacon) a series of very short
pulses towards the observer.
Consider the signal pulse emitted at a time tE in the reference frame S.

(b) Draw on the spacetime diagram of part (a) the worldline of the beacon. Hence show
that this emission event has the spacetime cordinates [tex] (ct_E,x_E)=(ct_E,c^2/a)[/tex]
(c) Show that the spacetime displacement of this emission event with respect to the origin
of S is given by the four vector
[tex]
\textbf{s}_E=ct_E\textbf{e}_0+(c^2/a)\textbf{e}_1
[/tex]
(d) Draw on the spacetime diagram the world line of the pulse emitted at time tE:

Homework Equations


[tex]
(\cosh(x))^2-(\sinh(x))^2=1
[/tex]

A four vector is defined as:
[tex]
\textbf{s}=\Delta ct\textbf{e}_0+\Delta x\textbf{e}_1+\Delta y\textbf{e}_2+\Delta z\textbf{e}_3
[/tex]

The Attempt at a Solution


(a)
Rearranging the given equations:
[tex]\sinh(a\tau/c)=at/c[/tex][tex]\cosh(a\tau/c)=ax/c^2[/tex]
Substituting into the relation between sinh and cosh:
[tex](ax/c^2)^2-(at/c)^2=1[/tex][tex](a^2x^2-a^2c^2t^2)/c^4=1[/tex][tex]x^2=c^2t^2+c^4/a^2[/tex] (1)

(b)
when tau=0:
[tex]t(0)=0[/tex][tex]x(0)=c^2/a[/tex]
this part I'm not so sure of, but here is what I did:
using the equation of motion [tex]x_f=x_i+ut+(1/2)at^2[/tex][tex]x = -½ac^2t^2 + c^2/a[/tex]
rearranging:
[tex]ct=\sqrt{(c^2/a-x)/(a/2)}[/tex] (2)

I'm not quite sure of my solutions for (a) and (b). Assuming I have the correct equations (1) and (2), all that remains is to plot them on the same plot of ct vs x?

(c)
substituting the spacetime coordinates [tex](ct_E,c^2/a)[/tex] into the definition of a four-vector:
[tex]
\textbf{s}_E=(ct_E-0)\textbf{e}_0+(c^2/a-0)\textbf{e}_1+(0-0)\textbf{e}_2+(0-0)\textbf{e}_3=ct_E\textbf{e}_0+c^2/a\textbf{e}_1
[/tex]

(d)
For this part I'm assuming that I have to draw a straight line of slope 1 or -1, intersecting the wordline of the beacon at an arbitrary time coordinate t_E, such that it also intersects the wordline of the observer. Is this correct?

Any help would be appreciated :)
 
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  • #2
alc95 said:
(a) Rearranging the given equations:
[tex]\sinh(a\tau/c)=at/c[/tex][tex]\cosh(a\tau/c)=ax/c^2[/tex]
Substituting into the relation between sinh and cosh:
[tex](ax/c^2)^2-(at/c)^2=1[/tex][tex](a^2x^2-a^2c^2t^2)/c^4=1[/tex][tex]x^2=c^2t^2+c^4/a^2[/tex]
This is fine.

(b) when tau=0:
[tex]t(0)=0[/tex][tex]x(0)=c^2/a[/tex]
this part I'm not so sure of, but here is what I did:
using the equation of motion [tex]x_f=x_i+ut+(1/2)at^2[/tex][tex]x = -½ac^2t^2 + c^2/a[/tex]
rearranging:
[tex]ct=\sqrt{(c^2/a-x)/(a/2)}[/tex]
First, why do you expect the beacon to accelerate once it's released from the ship, because that's what you're assuming here? Second, what's ##u##? Did you set it to 0? Third, are you sure the Newtonian kinematic equations are valid here? After all, this is a special relativity problem.

I'd start by finding the velocity of the beacon when it's released.

Assuming I have the correct equations (1) and (2), all that remains is to plot them on the same plot of ct vs x?
Yes.

(d)
For this part I'm assuming that I have to draw a straight line of slope 1 or -1, intersecting the wordline of the beacon at an arbitrary time coordinate t_E, such that it also intersects the wordline of the observer. Is this correct?
The slope will be ##\pm 1##, but the line may or may not intersect with the world line of the observer.
 
  • #3
Thanks for your reply :)

Here is what I have come up with for part (b):
I have already worked out the coordinates for the release of the beacon as (x,ct)=(c^2/a,0). So the worldline of the beacon intersects that of the ship at that point. Now, as you pointed out, the beacon would not necessarily accelerate upon release. So, would it be a straight line that is tangent to the point of release on the ship's wordline? My reasoning for this is that since it is at rest relative to the ship at that instant, their velocities would be equal at that point when plotted on the spacetime diagram.

I'm fairly sure that my part (c) is correct. Could you confirm this?
 
  • #4
alc95 said:
Here is what I have come up with for part (b):
I have already worked out the coordinates for the release of the beacon as (x,ct)=(c^2/a,0). So the worldline of the beacon intersects that of the ship at that point. Now, as you pointed out, the beacon would not necessarily accelerate upon release. So, would it be a straight line that is tangent to the point of release on the ship's wordline? My reasoning for this is that since it is at rest relative to the ship at that instant, their velocities would be equal at that point when plotted on the spacetime diagram.
Sounds good.

I'm fairly sure that my part (c) is correct. Could you confirm this?
Looks good too.
 
  • #5
For the spacetime diagram in Q1, I got a wordline for the beacon as a straight vertical line, intersecting that of ship at x=c^2/a. I'm not quite sure about this.

Problem statement
Q1
(e) The pulse is received by the observer at the observer’s proper time R. Show that this
event can be represented, with respect to the origin of S, by the position four vector
[tex]
s_R=(c^2/a)(\sinh{(a\tau_R/c)}\textbf{e}_0+\cosh{(a\tau_R/c)}\textbf{e}_1)
[/tex]

(f) The world line of the pulse (consisting of zero-rest mass particles) can then be parameterised
in terms of the affine parameter as
[tex]
s(\lambda)=s_E (1-\lambda)+s_R\lambda
[/tex]
Write down the four-velocity u of this pulse.

(g) Using the fact that u is a null vector, show that u0 = u1.

(h) Using the result of part (g), show that the time of reception of the pulse will be given
by:
[tex]
\tau_R=-(c/a)\ln{(1-at_E/c)}
[/tex]

Attempt at a solution
Q1
(e)
I had no issues with this part.

(f)
subbing in s_E and s_R and evaluating, assuming my algebra is correct:
[tex]
s(\lambda)=(ct_E(1-\lambda)+\lambda(c^2/a)\sinh{(a\tau_R/c)})\textbf{e}_0+((c^2/a)(1-\lambda)+\lambda(c^2/a)\cosh{(a\tau_R/c)}))\textbf{e}_1
[/tex]
then, the four-velocity is, by differentiating:
[tex]
\textbf{u}=(\lambda c)(\cosh{(a\tau_R/c)}\textbf{e}_0+\sinh{(a\tau_R/c)}\textbf{e}_1)
[/tex]

(g)
for a null vector the absolute value should be 0:
[tex]
\sqrt{(u_0)^2+(u_1)^2}=0
[/tex]
[tex]
\pm((u_0)^2+(u_1)^2)=0
[/tex]

(h) I am not sure how to use the result of part (g) when solving for tau_R. Is the idea to substitute this into the four-vector derived in part (f)?
 

What is a worldline?

A worldline is a path traced by an object in four-dimensional spacetime, showing its position and movement over time.

How are worldlines and spacetime diagrams related?

Spacetime diagrams are visual representations of worldlines, typically using a two-dimensional plane to depict the three spatial dimensions and one dimension for time.

What is the significance of a worldline?

Worldlines are significant in understanding the concept of causality in spacetime. They show the relationship between events and how they are influenced by the curvature of spacetime.

Can a worldline be curved?

Yes, a worldline can be curved in the presence of a gravitational field. This is known as geodesic curvature and is a result of the curvature of spacetime.

How do worldlines and spacetime diagrams relate to Einstein's theory of relativity?

Einstein's theory of relativity states that the laws of physics are the same for all observers in uniform motion. Worldlines and spacetime diagrams help to visualize this concept by showing how the movement of objects is relative to different frames of reference in spacetime.

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