Spacetime intervals again - still figuring out the formula

[JesseM]

Now, I think I understand that, but I still don't see why we're substracting. I don't even see the link between that and your explenations. Could we start over from where I asked 'why substract D^2 from c^2dt^2?' ?

Did you read my earlier post #14 where I derived the formula d\tau^2 = dt^2 - (1/c^2)(dx^2 + dy^2 + dz^2) from the formula for time dilation in relativity, d\tau = dt/\gamma? If you substitute d\tau^2 = ds^2/c^2 into that you get ds^2 = c^2dt^2 - dx^2 - dy^2 - dz^2.

 

[NanakiXIII]

Besides the fact that I only understand half of the mathemathics you used (mainly step three and four of those four lines of latex are a bit blurry), I'm really looking for a more concrete explenation. Surely there's some logic behind it that one could understand?

 

[Mortimer]

Well, how about this: think of a rectangular triangle. If you calculate the length of one of the rectangular sides (A) from the hypotenuse (H) and the other rectangular side (B), you get: A^2=H^2-B^2
This reflects the Minkowski way of doing relativity.

The Euclidean way of doing relativity simply says: H^2=A^2+B^2
Both formulas essentially contain the same information. but the roles of the variables have switched.

In the Minkowski formula, A is the invariant while in the Euclidean formula H is the invariant. The latter requires all objects to have velocity c in 4D Euclidean space which will become clear from the examples I will give below:

Instead of
ds^2=d(ct)^2-dx^2-dy^2-dz^2 (1)
you can also say:
d(ct)^2=ds^2+dx^2+dy^2+dz^2 (2)
In (1), ds is the invariant. In (2), d(ct) is the invariant. We must now make plausible that we can switch this role just like that.

For this to work out, we must make the assumption that the vectorsum of velocities in all 4 dimensions always equals c. That means that the velocity in the time dimension must be:
v_{time}=\sqrt{c^2-v_{space}^2} (3)

For a photon ds^2=0 so equation (2) reduces to
d(ct)^2=dx^2+dy^2+dz^2,
which is simply true. dct is indeed Lorentz-invariant and the time-component vanishes in this case because the speed of the photon equals c.

A moving mass particle cannot reach lightspeed so ds^2>0.
The factor dct in equation (2) is now still Lorentz invariant as result of our assumption that the object moves with velocity c in 4D. After all, this is supposed to represent the 4-dimensional distance traveled in a time duration dt which should always equal c times dt.

ds in this case should then represent the distance traveled in the time dimension in a timeduration dt. N.B: this is not dt but v_{time}dt. So the total distance traveled is
\sqrt{(v_{time}dt)^2+dx^2+dy^2+dz^2)} <br /> = \sqrt{dt^2(\sqrt{c^2-v_{space}^2})^2+(v_{space}dt)^2}
= dct

So the Euclidean way does make sense as long as the assumption about the velocity c in 4D is maintained. And it gets rid of the minus sign that you have been struggling with all the time! Bottom line is that this minus sign is merely a result of the awkward way of doing relativity with the Minkowski model of space-time.

 

[JesseM]

Besides the fact that I only understand half of the mathemathics you used (mainly step three and four of those four lines of latex are a bit blurry)

You mean d\tau = dt/\gamma and d\tau^2 = ds^2/c^2? The first is just the standard formula for time dilation in relativity, it means that if I see a clock traveling at velocity v for some period of time dt in my frame (say, 1 hour), then the clock itself will appear slowed down in my frame, and I will only observe it to tick forward by d\tau = dt/\gamma = dt\sqrt{1 - v^2/c^2} during that time period. This time dilation formula can itself be derived from the requirement that light must be measured to travel at the same speed in all reference frames--see the "light clock" explanation http://www.kineticbooks.com/physics/17467/17486/sp.html .

The second formula should just be thought of as the definition of ds in terms of the proper time d\tau. As far as I know ds doesn't have any independent physical meaning, I think people just came up with ds because they wanted an invariant measure of "distance" in spacetime that had units of length, instead of units of time. Since the proper time is a frame-invariant type of "distance" between two events in spacetime that has units of time, multiplying it by a constant with units of length/time will give a frame-invariant measure of distance with units of length.

I'm really looking for a more concrete explenation. Surely there's some logic behind it that one could understand?

Well, I think the light-clock explanation gives a pretty concrete explanation for the relation between time experienced in my frame and the proper time elapsed on a clock that's moving in my frame, and deriving d\tau^2 = dt^2 - (1/c^2)(dx^2 + dy^2 + dz^2) from that is pretty straightforward. Also, once you understand the meaning of proper time, it's clear that it must be a frame-invariant quantity, since different frames can't disagree on how much time elapses on a physical clock as it passes two given points in spacetime (with 'points in spacetime' marked by particular physical events, like 'clock departs earth' and 'clock arrives at mars'). But if that's not concrete enough I don't know of any simpler explanation; I suppose you could just trust that the formula follows from the Lorentz transformation, and the Lorentz transformation follows from the assumption that the laws of physics should work the same way in all reference frames and the assumption speed of light should be the same in all reference frames.

 

[JesseM]

Instead of
ds^2=d(ct)^2-dx^2-dy^2-dz^2 (1)
you can also say:
d(ct)^2=ds^2+dx^2+dy^2+dz^2 (2)
In (1), ds is the invariant. In (2), d(ct) is the invariant. We must now make plausible that we can switch this role just like that.

I don't see how dt could be an invariant unless you change the definition of dt so it's no longer what's measured on physical clocks. Surely if you pick two physical events and ask observers in different frames to measure the time between them using their own clocks, they will not all get the same answer.

For this to work out, we must make the assumption that the vectorsum of velocities in all 4 dimensions always equals c.

How do you define "velocity" in the time dimension? What physical procedure would you use to measure it?

 

[Mortimer]

JesseM said:
I don't see how dt could be an invariant unless you change the definition of dt so it's no longer what's measured on physical clocks. Surely if you pick two physical events and ask observers in different frames to measure the time between them using their own clocks, they will not all get the same answer.

Perhaps its easier to see if written like this:
c^2=(ds/dt)^2+(dx/dt)^2+(dy/dt)^2+(dz/dt)^2
Since ds=cd\tau
this equals:
c^2=(cd\tau/dt)^2+v_{space}^2 (1)

cd\tau/dt is the same as v_{time}
which follows from the relations
v_{time}=\sqrt{c^2-v_{space}^2}=c/\gamma
and
d\tau=dt/\gamma

The c^2 in (1) is invariant by assumption.
As to your question how to measure the velocity in the time dimension, you see here one possible physical procedure but I prefer another one which is explained http://www.rfjvanlinden171.freeler.nl/dimensionshtml/node2.html. (that saves me a lot of typing in this post)
That text essentially explains that the expression:
cd\tau/dt
should actually be written as:
c\frac{d\tau/dx_5}{dt/dx_5}
which is the ratio between the proper-time-velocity and the time-velocity in a 5D space and generally defines the velocity in the time dimension as
cd\tau/dx_5
(because dt/dx_5=1 always)

t is the time as measured by an observer in his own rest frame.

 

[jdavel]

NanakiXIII,

Try this.

A theory of relativity gives the relation between measurements made in each of two frames of reference S and S', where S' is moving at a speed v with respect to S.

Specifically, it relates the results of measurements made in each frame of the distance between events (space intervals). And it relates the results of measurements made in each frame of the elapsed time between events (time intervals).

If the measured value for an interval (time or space) is the same in both frames, it's called absolute. If not, it's called relative. Speed is called absolute if the ratio of the space interval to the time interval (for which the speed is defined) is the same in both frames. If not, it's called relative.

According to classical relativity (what Galileo, Newton and everyone else believed until 1905):

1) Time intervals are absolute
2) Space intervals (in general) are relative
As a result of 1) and 2),
3) All measurements of speed (including the speed of light) must be relative.

According to Einstein's relativity (what everyone has believed since 1905):

1) Measurements of the speed of light are absolute.
2) Measurements of any speed less than the speed of light are relative
As a result of 1) and 2)
3) Time intervals AND space intervals must be relative.
As a result of the specific way that time intervals and space intervals are relative,
4) The time interval squared times the speed of light squared minus the space interval squared is absolute.

That's the logic of why the space time interval is what it is. To see mathematically why the spacetime interval is what it is, you have to do the math.

 

[JesseM]

Perhaps its easier to see if written like this:
c^2=(ds/dt)^2+(dx/dt)^2+(dy/dt)^2+(dz/dt)^2
Since ds=cd\tau
this equals:
c^2=(cd\tau/dt)^2+v_{space}^2 (1)

cd\tau/dt is the same as v_{time}
which follows from the relations
v_{time}=\sqrt{c^2-v_{space}^2}=c/\gamma
and
d\tau=dt/\gamma

The c^2 in (1) is invariant by assumption.

OK, I agree that if you choose to define "velocity through time" as (cd\tau/dt) then this math works out. But your equation only has c^2 as an invariant, not dt; why did you earlier say that "In (2), d(ct) is the invariant"?

As to your question how to measure the velocity in the time dimension, you see here one possible physical procedure but I prefer another one which is explained http://www.rfjvanlinden171.freeler.nl/dimensionshtml/node2.html. (that saves me a lot of typing in this post)
That text essentially explains that the expression:
cd\tau/dt
should actually be written as:
c\frac{d\tau/dx_5}{dt/dx_5}
which is the ratio between the proper-time-velocity and the time-velocity in a 5D space and generally defines the velocity in the time dimension as
cd\tau/dx_5
(because dt/dx_5=1 always)

t is the time as measured by an observer in his own rest frame.

If you define "velocity through time" as (cd\tau/dt) then of course all you have to do to measure it is to measure both the elapsed time in your reference frame and the elapsed proper time on the moving clock, and plug them into that formula. But above you seem to suggest that your concept of "velocity through time" is not just a definition in terms of existing ideas in relativity, but is actual part of some new theory you have devised which involves an extra dimension (or two extra dimensions?) beyond the three space and one time dimensions we are accustomed to. If you're discussing a new theory of your own, then that really shouldn't go in this forum--as said in the IMPORTANT! Read before posting sticky thread at the top of this forum, this forum is just meant for discussing the theory of relativity, and "is not meant as a soapbox for those who wish to argue Relativity's validity, or advertise their own personal theories". If you want to discuss a new theory of your own, that's what the https://www.physicsforums.com/forumdisplay.php?f=12 forum is for.

 

[PeteSF]

Another thing to notice is that the spacetime interval defined above is just c times the proper time (time as measured on the clock of someone moving along that worldline), d\tau^2 = dt^2 - (1/c^2)(dx^2 + dy^2 + dz^2)

Since all frames must agree on how much time ticks on a clock between two points on its worldline, this shows why it's necessary that d\tau, and hence ds = cd\tau, must be a frame-invariant quantity.

Thinking about spacelike intervals...

The interval would be the proper length of a rod with position and (inertial) motion such that the two ends of the rod coincided with the two events in question simultaneously in the rod's rest frame...

Right? And is there a simpler way of saying it?

 

[JesseM]

Thinking about spacelike intervals...

The interval would be the proper length of a rod with position and (inertial) motion such that the two ends of the rod coincided with the two events in question simultaneously in the rod's rest frame...

Right? And is there a simpler way of saying it?

That can't be right, because the separation between two events which are simultaneous in some inertial frame must be spacelike, and both ds and d\tau are imaginary for spacelike intervals.

 

[PeteSF]

You're right... My understanding grows a little, and I'd better do the maths.

***

OK, it seems the interval is i times the proper length:

\Delta s^2 = c.\Delta t^2 - (\Delta x^2 + \Delta y^2 + \Delta z^2)
\Delta s = i\sqrt{\Delta x^2 + \Delta y^2 + \Delta z^2}

Is this result just a fun fact, or does it lead to any insights?

 

[Mortimer]

JesseM said:
But above you seem to suggest that your concept of "velocity through time"
is not just a definition in terms of existing ideas in relativity, but is
actual part of some new theory you have devised which involves an extra
dimension (or two extra dimensions?) beyond the three space and one time
dimensions we are accustomed to. If you're discussing a new theory of your
own, then that really shouldn't go in this forum

Point taken. My fault. In this thread I'll stick to the definition cd\tau/dt for the velocity in the time dimension.

OK, I agree that if you choose to define "velocity through time" as cd\tau/dt then this math works out. But your equation only has c^2 as an invariant, not dt; why did you earlier say that "In (2), d(ct)^2 is the invariant"?

For the same reason as d(c\tau)^2 is the invariant in the Minkowski style of this equation.
When I rewrote the Minkowski equation
ds^2=d(ct)^2-dx^2-dy^2-dz^2 (1)
into the Euclidean equation
d(ct)^2=ds^2+dx^2+dy^2+dz^2, (2)
I switched the role of the invariant with the time component in the 4-vector. Indirectly this also switches the role of t and \tau (because ds=cd\tau). The switch was mathematically justified by the assumption of the universal velocity c for objects in 4D space-time.
Equations (1) and (2) describe the Minkowski and Euclidean 4-vectors for position.
Equation (2) is transformed into the Euclidean velocity 4-vector
c^2=(ds/dt)^2+(dx/dt)^2+(dy/dt)^2+(dz/dt)^2
by differentiation with respect to t (which is what we are supposed to do if we calculate velocities).
The Minkowski 4-vector for velocity is derived from the 4-vector for position by differentiation with respect to \tau instead of t. The reason for this is historically determined: differentiation with respect to t results in the 4-vector (c, v_1, v_2, v_3) but this 4-vector is not invariant under a Lorentz transformation in Minkowski space. Therefor, the 4-vector was multiplied with the factor \gamma to artificially make it invariant. This \gamma leads to the transformation of t into \tau. The Euclidean 4-vector is invariant as is but requires the assumption of the universal velocity c in 4D space-time.

As I indicated earlier in this thread, Euclidean relativity is not new. It is at least 40 years old. A very clean description (i.e., restricting itself to the core) is published by prof. Alexander Gersten (Ben-Gurion univ., Israel) and can be found here: www.bgu.ac.il/~gersten/papers/euclirev.pdf
Other articles on Euclidean relativity (including mine) often stack a number of speculations on top of it.

 

[JesseM]

You're right... My understanding grows a little, and I'd better do the maths.

***

OK, it seems the interval is i times the proper length:

\Delta s^2 = c.\Delta t^2 - (\Delta x^2 + \Delta y^2 + \Delta z^2)
\Delta s = i\sqrt{\Delta x^2 + \Delta y^2 + \Delta z^2}

Is this result just a fun fact, or does it lead to any insights?

Yeah, that works, I hadn't thought of it like that before. It'd be even simpler if you forgot about ds and instead used dS, defined like this:

dS^2 = dx^2 + dy^2 + dz^2 - c^2 dt^2

Then for two events with a spacelike separation, dS would simply represent the distance between the events as measured in the frame where they happened simultaneously. Obviously in that frame, dt=0, so dS is just the ordinary distance formula. But this seems more like a "fun fact" rather than something that leads to any big insights.

 

[PeteSF]

Thanks Jesse!

 

[Ich]

But this seems more like a "fun fact" rather than something that leads to any big insights.

dS^2 is interpreted either as a proper time or as a proper length, depending on the sign. Both are equally valid, so I don´t see why one of those interpretations should be a fun fact.

 

[Mortimer]

I agree with "Ich". I can see what JesseM tries to say but its a bit confusing to use the term "space-like" in this context as well as the expression dx^2+dy^2+dz^2-c^2dt^2, which is the interval using the -+++ metric instead of the +--- metric.

 

[JesseM]

dS^2 is interpreted either as a proper time or as a proper length, depending on the sign.

What do you mean by "depending on the sign"? You have to change more than just the sign to go from the proper length dS between events with a spacelike separation to the proper time d\tau between events with a timelike separation:

dS^2 = dx^2 + dy^2 + dz^2 - c^2 dt^2

d\tau^2 = dt^2 - (1/c^2)(dx^2 + dy^2 + dz^2)

So, dS = i*c*d\tau

Both are equally valid, so I don´t see why one of those interpretations should be a fun fact.

I agree both are equally valid, I just meant that if you already understand how to get the equation for d\tau, you don't really gain any big new insight by figuring out the derivation for dS...but the reverse is true as well I suppose (although the expression for proper time is useful in more problems since you can integrate d\tau over a curved path to find the total proper time, while I don't think there's any situation where you'd want to integrate dS).

 

[JesseM]

I agree with "Ich". I can see what JesseM tries to say but its a bit confusing to use the term "space-like" in this context

Why is it confusing? The concept of the distance between two events in the frame where they are simultaneous only makes sense if there is a spacelike separation between them, if they are timelike separated there is no frame where they are simultaneous.

as well as the expression dx^2+dy^2+dz^2-c^2dt^2, which is the interval using the -+++ metric instead of the +--- metric.

Again, if you want to find the distance between two events in the frame where they are simultaneous, you have to write it that way. If you used c^2 dt^2 - dx^2 - dy^2 - dz^2, that would give i times the distance.

 

[Mortimer]

No big deal. Strictly speaking you are not saying anything wrong.
I see no physical difference between the +--- metric and the -+++ metric. Although there is a mathematical difference (the i) I find this hardly relevant because the Minkowski "distance" between events is not a real distance in the sense as we use it in space anyway.

 

[NanakiXIII]

JesseM, no, I meant those four lines you used to get to the final formula.

3) Time intervals AND space intervals must be relative.
As a result of the specific way that time intervals and space intervals are relative,
4) The time interval squared times the speed of light squared minus the space interval squared is absolute.

That's the logic of why the space time interval is what it is. To see mathematically why the spacetime interval is what it is, you have to do the math.

I'm going to assume 3 is correct, I don't fully understand. But I don't see how 3 leads to 4.


My apologies for being away for a bit, I've been and am ill. Also, sorry for not replying to all your replies, I just picked out what seemed important.

 

[PeteSF]

I find this hardly relevant because the Minkowski "distance" between events is not a real distance in the sense as we use it in space anyway.

...but it is for events with spacelike separation, yes?

 

[Mortimer]

The point is that with a spacelike separation the invariant ds becomes complex (because ds^2&lt;0). If you don't like that you can multiply the whole interval with i and get a real number again. That's really just a matter of convention. But then again, this whole situation is purely academic anyway because a spacelike distance between events can only be demonstrated at velocities > c which is still considered an impossibility.
For your understanding, it is probably important to remember that the distance between events (i.e. ds) in Minkowski space-time is just a "gadget" that happens to be invariant under a Lorentz transformation. It has no real physical meaning like for instance a distance in space or a timelapse on a clock. The math is merely a tool that helps making the correct calculations (just as complex numbers are for instance).

 

[PeteSF]

But then again, this whole situation is purely academic anyway because a spacelike distance between events can only be demonstrated at velocities > c which is still considered an impossibility.

What?
Any pair of events that are simultaneous in some reference frame have a spacelike separation, yes?
FTL speeds are required for information transfer between such events, but is that relevant?

 

[Mortimer]

Any pair of events that are simultaneous in some reference frame have a spacelike separation, yes? FTL speeds are required for information transfer between such events

Yes. So there is no way that you can ever "see" or "measure" such a situation (note that I deliberately said "demonstrated", not "exist").

but is that relevant

Well, depends on what you want to do with it, doesn't it? If i will never ever be able to come across the situation, what's the point in discussing it like it is real? It's like speculating about tachyons and so.

 

[Ich]

Yes. So there is no way that you can ever "see" or "measure" such a situation (note that I deliberately said "demonstrated", not "exist").

Spacetime has to be measured with RODS and CLOCKS, according to Einstein.
Either a length, measured (at least symbolically) with rods, or a time difference, measured with clocks, are physically meaningful and well defined. There exists a prescription how to measure proper length, and it is invariant. What more do you want?

But JesseM is right, as world lines are necessarily timelike, ds=dtau will be used more often in calculations and is therefore more useful in a way. But it´s not the whole truth about the spacetime interval. You need both interbpretations posslible to be consistent.

 

[robphy]

Measuring displacements in Minkowski spacetime

But then again, this whole situation is purely academic anyway because a spacelike distance between events can only be demonstrated at velocities > c which is still considered an impossibility.
For your understanding, it is probably important to remember that the distance between events (i.e. ) in Minkowski space-time is just a "gadget" that happens to be invariant under a Lorentz transformation. It has no real physical meaning like for instance a distance in space or a timelapse on a clock. The math is merely a tool that helps making the correct calculations (just as complex numbers are for instance).

Any pair of events that are simultaneous in some reference frame have a spacelike separation, yes? FTL speeds are required for information transfer between such events

Yes. So there is no way that you can ever "see" or "measure" such a situation (note that I deliberately said "demonstrated", not "exist").


Well, depends on what you want to do with it, doesn't it? If i will never ever be able to come across the situation, what's the point in discussing it like it is real? It's like speculating about tachyons and so.

This may be helpful.


Here's a procedure to measure displacements in [Minkowski] spacetime using a clock, a light source, and detector. (This is the "radar method".)
Given two events, P and Q, what is the square-interval?

Consider an inertial observer meeting P.
One his worldline, there is
an event S when at a light ray can be sent to Q, and
an event R when that light ray's reflection is received from Q.

Let t_P, t_S, t_R be the times read off that observer's clock. (Obviously, t_R \geq t_S.)

According to this observer, the spatial-displacement from P to Q is
\Delta x_{Q\mbox{ from }P}=c(t_R-t_S)/2 ,
that is, half of the measured round-trip time multiplied by the speed of light.

According to this observer, the time-coordinate of the distant-event Q
is DEFINED by t_Q= (t_R+t_S)/2,
that is, the average of the send and receive clock-readings.

So,
according to this observer, the time-displacement from P to Q
is \Delta t_{Q\mbox{ from }P}=t_Q-t_P,
or
\Delta t_{Q\mbox{ from }P}=(t_R+t_S )/2 -t_P.
that is, the average of the send and receive clock-readings minus the clock-reading t_P.

The following quantity can tell us about the causal relationship of Q from P.
( t_R - t_P)(t_S - t_P).

If the events on this worldline happen in the sequence S-then-P-then-R,
then Q is spacelike-related to P. This quantity is negative.

If the events on this worldline happen in the sequence P-then-S-then-R,
then Q is in the timelike-future of P. This quantity is positive (since it's the product of two positive numbers).

If the events on this worldline happen in the sequence S-then-R-then-P,
then Q is in the timelike-past of P. This quantity is positive (since it's the product of two negative numbers).

If any two events coincide, then that quantity is zero.
If it's S and P that coincide, then Q is in the lightlike-future of P.
If it's R and P that coincide, then Q is in the lightlike-past of P.
If S, R, and P coincide, then Q coincides with P.


Now, consider any two inertial observers that meet P and
perform this procedure to make measurements of Q.
So, each observer will have a different set of send and receive events and
thus a different set of clock-readings for send and for receive.

According to special relativity, the quantity
c^2( t_{receive} - t_P)( t_{send} - t_P) is invariant.
Indeed,
<br /> \begin{align*}<br /> (c\Delta t_{Q\mbox{ from }P})^2-(\Delta x_{Q\mbox{ from }P})^2<br /> &amp;=c^2\left(\frac{t_R+t_S}{2} -t_P\right)^2-\left(c\frac{t_R-t_S}{2} \right)^2\\<br /> &amp;=c^2\left(\frac{(t_R-t_P)+(t_S-t_P)}{2}\right)^2-\left(c\frac{(t_R-t_P)-(t_S-t_P)}{2} \right)^2\\<br /> &amp;=c^2\left( \left(\frac{(t_R-t_P)+(t_S-t_P)}{2}\right)^2-\left(\frac{(t_R-t_P)-(t_S-t_P)}{2} \right)^2\right) \\<br /> &amp;=c^2\left( \frac{2(t_R-t_P)(t_S-t_P)}{4} - \frac{-2(t_R-t_P)(t_S-t_P)}{4} \right) \\<br /> &amp;=c^2 (t_R-t_P)(t_S-t_P) \\<br /> \end{align*}<br />
(This calculation is simpler if we assume t_P=0.
To keep to the spirit of emphasizing the time-measurements, one should start at the bottom with (t_R-t_P)(t_S-t_P) and obtain the expression (c\Delta t)^2- \Delta x^2.)


So, to comment on the sections I quoted above,
there is a way to measure (with a physical setup) the separation of two spacelike-related events.

 

[JesseM]

JesseM, no, I meant those four lines you used to get to the final formula.

OK, if you're comfortable with these formulas:

Time dilation:
d\tau = dt/\gamma = dt\sqrt{1 - v^2/c^2} (1)

Velocity = distance/time:
v = \sqrt{dx^2 + dy^2 + dz^2}/dt (2)

Then those last four lines are just a matter of algebra:

First, start with equation (1):
d\tau = dt \sqrt{1 - v^2/c^2}

Then square both sides of equation (2) and substitute in for v^2:
d\tau = dt \sqrt{1 - (dx^2 + dy^2 + dz^2)/(c^2dt^2)}

Then factor out 1/dt^2 from the expression inside the square root:
d\tau = dt \sqrt{(1/dt^2)(dt^2 - (1/c^2)(dx^2 + dy^2 + dz^2))}

Then do \sqrt{(1/dt^2)*stuff} = (1/dt)\sqrt{stuff}, and the 1/dt cancels out with the dt that was already outside the square root:
d\tau = \sqrt{dt^2 - (1/c^2)(dx^2 + dy^2 + dz^2)}

If there's still any of this algebra you're not clear on, let me know which step is giving you trouble.

 

[NanakiXIII]

I don't understand how you got to the third formula

d\tau = dt \sqrt{(1/dt^2)(dt^2 - (1/c^2)(dx^2 + dy^2 + dz^2))}

 

[jtbell]

It may be a little bit more obvious if we first rewrite the preceding equation in a slightly different format:

d\tau = dt \sqrt{1 - \frac {dx^2 + dy^2 + dz^2}{c^2dt^2}}

Now let's proceed:

d\tau = dt \sqrt{\frac {dt^2}{dt^2} - \frac {dx^2 + dy^2 + dz^2}{c^2dt^2}}

d\tau = dt \sqrt{\frac {1}{dt^2} dt^2 - \frac {1}{dt^2} \frac {dx^2 + dy^2 + dz^2}{c^2}}

d\tau = dt \sqrt{\frac {1}{dt^2} \left( dt^2 - \frac {dx^2 + dy^2 + dz^2}{c^2}\right)}

 

[JesseM]

I don't understand how you got to the third formula

d\tau = dt \sqrt{(1/dt^2)(dt^2 - (1/c^2)(dx^2 + dy^2 + dz^2))}

Well, do you understand how to factor variables out of equations in algebra? For example, do you understand why if you factor x^2 out of (x^3 + 2x^5 + 5) you'd get (x^2)*(x + 2x^3 + 5/x^2)? If so, just factor (1/dt^2) out of the equation (1 - (1/dt^2*c^2)(dx^2 + dy^2 + dz^2)) in the same way.