# I What is ds in ds2 = dx2 - cdt2

#### Pedroski55

Summary
Puzzled by the apparent outcome of the formula.
Hi,
I'm just an amateur, no physics qualifications. I've been reading here.

Professor Richard Conn Henry likes the word trivial. None of this is trivial for me.

Einstein apparently said:
(sorry don't know how to present the maths correctly here, dx2 means 'dx squared'.)

dx2 + dy2 + dz2 - dt2 = ds2 = dx'2 + dy'2 + dz'2 -dt'2 (i.e. ds is invariant for any observer)

Professor Henry goes quickly on to show how photons know no time. I'm a bit (lot) slower than him and light.

So I'm just looking at:
(assume, as they do, the motion is only in x)

ds2 = dx2 - dt2

The German word for distance is Strecke. So German mathematicians will often use S or s for Strecke, distance, in formulas. I assume, ds is a distance, the distance from the point of origin, in whatever units, let's say metres, as are dy and dz.

Now, as light seems to have a constant speed, we can regard 1 second of time as "the length light travels in 1 second". So multiply time by, let's say c = 300 000 metres per second, as a round number, and we have a 'light distance second' in units of metres. I think that is what they are doing. Certainly, you can't just add seconds to metres.

ds2 = dx2 - cdt2

In Einstein's 4-dimensional Pythagorean type calculations I get some funny results.

Assume the velocity of the object I observe is 1 metre per second and time, dt, is 1 second, so I observe something travel 1m, dx = 1m

ds2 = 1 - 300 000 * 1 = - 299 999 metres, ds = √-299 999

Actually, I observe the object moved 1m from me in 1s and therefore ds = 1 = dx

Obviously, I don't know what I'm doing. Could you please point me right? What exactly is Einstein's ds?

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#### PeroK

Homework Helper
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2018 Award
Actually, I observe the object moved 1m from me in 1s and therefore ds = 1 = dx

Obviously, I don't know what I'm doing. Could you please point me right? What exactly is Einstein's ds?
In normal Euclidean geometry $ds^2$ is never negative, so you do have $ds = \sqrt{ds^2}$.

But, in spacetime geometry $ds^2$ can be negative, as you have found out. In this case the length of the spacetime interval is a time, often denoted by $d\tau$. And, you have, by definition:

$d\tau = \sqrt{-ds^2}$

• Pedroski55, lomidrevo and Ibix

#### Ibix

First off, it's very common in relativity to use units such that $c=1$, such as seconds for time and light seconds for distance. It just saves typing. You seem to be understanding this almost correctly. Where you've gone wrong is that you wrote $c$ instead or $c^2$, but you can write $ds^2=dx^2-c^2dt^2$ if you prefer.

Second, there's guidance on writing formulae in the forum at the "LaTeX guide" link below the reply box.

And to answer your main question - the quantity $ds$ is usually called interval. It's not distance travelled in space - it's distance travelled in spacetime. It turns out to be closely related to the elapsed time on the wristwatch of the person (or object - any clock attached to it will do, even if it doesn't have wrists) doing the travelling. And spacetime has the rather peculiar property that its equivalent of Pythagoras' Theorem has one negative sign in it. Where you appear to be going wrong is expecting that the "distance" travelled through spacetime will be the same as the distance traveled through space. It isn't.

One warning - there is no universal agreement on whether $ds^2=dt^2-(dx^2+dy^2+dz^2)$ or $ds^2=(dx^2+dy^2+dz^2)-dt^2$. Either convention is fine and makes no difference to anything, unless you carelessly mix up results from the two conventions.

So, your object that moves 1m through space in 1s follows a path through spacetime that has interval $\sqrt{(1 m)^2-(3\times 10^8 m/s\times 1 s)^2}$. Note also that you've used the speed of light in km/s, but are measuring distances in m - that won't work. I've used the correct numbers. As PeroK says, the interval turns out to be related to the elapsed time by $ds^2=-c^2d\tau^2$, so the time clock on the object will be very slightly behind one at the far end of your meter rule - it will show $\sqrt{1-1/(9\times 10^{16})}$s to have elapsed - roughly half a millionth of a nanosecond less.

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• Pedroski55 and PeroK

#### Pedroski55

Thank you both very much! Sorry about getting the speed of light wrong!

It is however hard for me to imagine a negative length and impossible for me to imagine spacetime!

The is I believe, but the square root of a negative number is unreal.

Today, I sat on a wall by the lake in the sunshine, closed my eyes and felt the warm sun on my face. People say it takes 8 minutes for light to reach us from the sun.

But for the photons, time does not exist, according to Professor Henry! as v --> c ds --> zero! Of course, if no time passes, no distance can be travelled.

What warms my face in the sunshine? (Just a rhetorical question, an answer is impossible I think!)

I read this below. He seems to be suggesting, the negative ds squared value could be a consequence of symmetry.

Nothing but Relativity, Palash B. Pal
Saha Institute of Nuclear Physics, 1/AF Bidhan-Nagar, Calcutta 700064, INDIA

Let us consider two inertial frames S and S′, where the second one moves with a speed v, along the x-axis, with respect to the first one. The co-ordinates and time in the S-frame will be denoted by x and t, and in the frame S′, they will be denoted with a prime. The space-time transformation equations have the form

x′=X(x, t, v),(1)

t′=T(x, t, v),(2)​

and our task is to determine these functions. A few properties of these functions can readily be observed. First, the principle of relativity tells us that if we invert these equations, we must obtain the same functional forms:

x=X(x′, t′,−v),(3)

t=T(x′, t′,−v).(4)​

Notice that here the third argument of the functions is−v, since that is the velocity of the frame S with respect to S′.

#### PeroK

Homework Helper
Gold Member
2018 Award
It is however hard for me to imagine a negative length and impossible for me to imagine spacetime!
That doesn't give you much chance to learn SR. Not all geometry is Euclidean.

In general, if you are not used to studying mathematics and physics, then you will need to free your mind. And, you need to ask yourself why you want to learn this. When you reach what you think may be a contradiction in physics, you have to ask yourself:

a) do professional physicists really not notice that - in this case - if time does not exist for a photon then a photon cannot travel anywhere?

b) maybe you have misunderstood something or are making a false assumption about what something means?

If you choose b), then you can start to learn some physics.

The answer to the question about photons is that there is no frame of reference that travels at the speed of light. So, you cannot study the universe from the point of view of a photon. But, you can study the universe from any other reference frame, in which photons always travel at the speed of light.

In terms of spacetime intervals, photons travel along "null" paths, where the spacetime interval is of zero length.

• Pedroski55, lomidrevo and Dale

#### haushofer

But for the photons, time does not exist, according to Professor Henry!
For something to "not exist" is something different than "it is zero". It's simply an ill-defined question to ask "how much time has passed for a photon". For that you have to compare your clock with a photon's, but that means you have to get into its restframe, so it's speed would become zero. But according to relativity, every inertial frame will measure a lightspeed of c = 3*10^8 m/s. So that's a contradiction.

• lomidrevo, PeroK and vanhees71

#### Dale

Mentor
What warms my face in the sunshine? (Just a rhetorical question, an answer is impossible I think!)
An answer is very possible, but the question belongs in a separate thread in the classical physics sub forum rather than here.

It is however hard for me to imagine a negative length and impossible for me to imagine spacetime!
That is where you should start then. What makes imagining spacetime difficult? Have you ever seen a graph of position vs time? That is spacetime.

• Pedroski55

#### Ibix

It is however hard for me to imagine a negative length and impossible for me to imagine spacetime!
I don't think it's possible to imagine. You just have to accept it. The consequences of the world being this way are fairly specific and well-matched to experiment.
But for the photons, time does not exist, according to Professor Henry!
You can't define elapsed time along a null worldline, it's true. That doesn't mean light can't travel - it just means that you can't describe it travelling in the way you naively think we ought to be able to. It clearly does travel.

The eight minutes is eight minutes measured by us - what the light itself does or does not measure (assuming that the concept of light measuring anything makes any sense) is irrelevant to that.
I read this below. He seems to be suggesting, the negative ds squared value could be a consequence of symmetry.
More or less all modern physics is the result of requiring physics not to change under certain transformations - or requiring certain symmetries. Pal's paper is rather neat, although I don't think it's the only attempt t what he does. He shows that the only two systems of physics consistent with the principle of relativity are Galilean relativity (i.e. Newtonian physics) and Einstein's relativity. Experiment tells us we are not in a Newtonian universe.

• Pedroski55

#### Mister T

Gold Member
It is however hard for me to imagine a negative length and impossible for me to imagine spacetime!
You need to begin with the concept of an event. Something that happens at a particular location, say $x_1$ and a particular time, say $t_1$. This event might be me snapping my fingers. Now consider a second event, at a location $x_2$ and a time $t_2$. It might be you snapping your fingers. The difference in positions is $\Delta x=x_2-x_1$ and the difference in clock readings is $\Delta t=t_2-t_1$. Let's measure $x$ and $t$ in the same unit, say minutes for time and minutes for distance (where a minute of distance is the distance light travels in one minute of time).

If a light beam leaves from my location when I snap my fingers and arrives at your location when you snap your fingers, then $\Delta x = \Delta t$ and we say the interval between the events is lightlike. On the other hand, if you snap your fingers before that light beam arrives then $\Delta x$ is larger than $\Delta t$ and we say the interval is spacelike. On the other hand, if you snap your fingers after that light beam arrives then $\Delta t$ is larger than $\Delta x$ and we say the interval is timelike.

For timelike intervals, the value of the interval equals the time that elapses for someone who is present at both events. Of course, it's not possible for someone to be present at both events if the interval is lightlike or spacelike.

The connection between the different types of intervals, simultaneity, and causality are revolutionary ideas. Of course, they've been well understood now for over a century.

• Pedroski55

#### Pedroski55

Thanks again everybody! I'll go sit on my little wall and wonder about all this some more!

"What is ds in ds2 = dx2 - cdt2"

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