4 ms·

In the sense that GP states, yes, but I disagree with GP: "E=m" is wrong even if c=1 in the units you're working with since c is not unitless; it has speed unit

In the sense that GP states, yes, but I disagree with GP: "E=m" is wrong even if c=1 in the units you're working with since c is not unitless; it has speed units. You can't have a kilogram of energy. You need to multiply a mass by something with squared-speed units to get something with energy units.

chriswarbo 9y ago
> "E=m" is wrong even if c=1 in the units you're working with since c is not unitless In natural units, c is "unitless" (AKA dimensionless), by definition.
- throwaway287391 9y ago
  I won't say you're wrong since I stopped taking physics in high school, but this makes absolutely no sense to me. How does defining the Planck speed unit to be the speed of light suddenly make c a unitless constant? It's not "c=1" (except in informal shorthand), it's "c = 1 Planck speed unit". It seems like all sorts of things would "break" very quickly when you start dropping units. Maybe someone has a link that goes into more depth on why all of this is okay?
- raattgift 9y ago
  The value of c is arbitrary and depends on one's choice of coordinates. However the presence of c in the line element of a spacetime like ours is mandatory. There is, however, some conceptual value in setting c to some value other than 1, which I'll return to in the last paragraph below. When using a pseudo-Riemannian manifold one uses a metric signature where (keeping it simple, cf. "metric signature" on wikipedia) coordinates on orthogonal dimensions of can take one sign, or the opposite sign and a constant multiplier. In the Lorentzian case there will be one dimension taking one sign, and one or more taking the other; the choice of whether the solitary dimension takes a + or - sign is a matter of convention or preference. Conventionally the solitary dimension also takes the constant multiplier and is called the timelike dimension. A Lorentzian signature (minuses, 1) or (1, plusses) guarantees that one can describe paths through the manifold as null, timelike, or spacelike, and this gives one a causal structure. Our universe can be well represented by a Lorentzian manifold (3, 1) or (1, 3), and this has been tested to exquisite precision. It does not tell us the value of the constant c, but we can determine that from tests of causal relations, the boundaries of which will be null. Alternatively, a massless pointlike object will always travel on null geodesics. One runs into c being set to unity in systems geometrized units in relativity often; it's very handy to have mark off coordinates as e.g. "-seconds" vs "c seconds" (-,+,+,+ aka (1,3)) as long as one doesn't mess up one's dimensional analyses (which is unfortunately easy). As a concrete example, the line element for (1,3) flat spacetime using Cartesian coordinates is dS^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2. Compare the formula for Euclidean distance in the (Euclidean flat) plane between points p = (p1,p2) and q = (q1,q2) for (x,y) coordinates: d(p,q) = sqrt((q1-p1)^2 + (q2-p2)^2), which we could rewrite as dx = q1-p1, dy = q2-p2, ds = sqrt(dx^2 + dy^2) or ds^2 = dx^2 + dy^2. In Euclidean 3-space, we add another axis: ds^2 = dx^2 + dy^2 + dz^2. In Lorentzian 4-spacetime, we have to change the sign, so ds^2 = dx^2 + dy^2 + dz^2 - c^2dt^2. Note that we are not restricted to use any particular unit of distance or system of units; dx could be in metres, miles, astronomical units, light-years, gigaparsecs or practically anything else, while dt could be in seconds or fortnights or any other handy unit of time. When setting c to unity, we do need to choose appropriate units for dx (and dy and dz ...) vs dt. In SI units, that's seconds and light-seconds, but we could use another system if we wanted. Notably we aren't restricted to Cartesian coordinates, however if we were to change to some other system of coordinates (e.g. polar ones) the line element would need to reflect that. For example, in the Lorentzian 4-spacetime case, we would write ds^2 = dr^2 + r^2 * dtheta^2 + r^2sin^2(theta)dphi^2 -c^2dt^2. Finally, when using SI units (for example), one can see very clearly that a path taken by an object moving much slower than the speed of light is totally dominated by the amount of time between starting point and finishing point, because the value of c is large. Using the (+,-,-,-) metric signature [ds^2 = c^2dt^2 - dx^2 - dy^2 - dz^2], a large c helps make it clear that lightlike paths through spacetime are shorter, and purely timelike paths (where dx=0, dy=0, dz=0, dt != 0) have extremized length, since we don't subtract anything from cdt. This is the root of the explanation of the twin paradox in flat spacetime: the twin moving quickly compared to the speed of light takes a shorter path between together1 = (x1,y1,z1,t1) and together2 = (x2,y2,z2,t1) than the twin moving slowly compared to the speed of light. In the extreme, twin A holds x=const,y=const,z=const, whereas twin B's x coordinate is only equal to twin A's at the start and end of the journey. This holds up under any system of coordinates; we could consider r=const vs changing r in spherical coordinates, for instance.
- BeetleB 9y ago
  I won't profess knowledge of particle physics, but in EM systems where they set c=1, c is not dimensionless.
mikeash 9y ago
Consider the more mundane example of heat and work. The standard unit of heat is the calorie, which is the heat needed to increase the temperature of one gram of water by one degree celsius under standard conditions. The standard unit of work is the joule, which is the work needed to exert a force of one newton over a distance of one meter. James Joule discovered that heat and work can be related by: h = kw Where k is a constant equal to 4.184 calories/joule. But really, this isn't a relationship between heat and work. This is a statement of equivalence. The constant of 4.184cal/J just tells you how the two units differ. They both describe the same property. Really, h = w. The version with k is only needed if you insist on using the traditional, different units on the left and right sides. The same is true of e = mc^2. This does not tell you how two different fundamental properties are related, it merely tells you how two units, traditionally considered separate, actually measure the same underlying property. Really, e = m. The c^2 is only needed if you insist on using J on one side and kg on the other. That is to say, you absolutely can have a kilogram of energy. If you have an object with a mass of one kilogram, you do have a kilogram of energy. Physicists and engineers don't typically use that unit for energy, but there's no reason they couldn't, and likewise no reason they couldn't use joules for mass.
- throwaway287391 9y ago
  Thank you for the nice explanation, despite my somewhat rude and uninformed posts! I had some trouble accepting it, but the heat/work example made it "click" for me, once my brain got past its initial stubborn reaction of "but of course heat and work are both energy, whereas mass is clearly not!" So any time two physical quantities are related by a constant factor -- even if said constant factor has what we currently think of as "units" (and even particularly weird units like squared speed) -- they are fundamentally equivalent. I'm still mulling this over, but it's a pretty fascinating mental revelation for me.
  - mikeash 9y ago
    Nice. I thought the heat/work thing was an interesting angle on it and I'm glad it worked. Edit: I also want to say that I didn't find the comment I replied to rude in any way. It's direct and straightforward, but that's pretty common around here.

chriswarbo 9y ago

> "E=m" is wrong even if c=1 in the units you're working with since c is not unitless In natural units, c is "unitless" (AKA dimensionless), by definition.

throwaway287391 9y ago
I won't say you're wrong since I stopped taking physics in high school, but this makes absolutely no sense to me. How does defining the Planck speed unit to be the speed of light suddenly make c a unitless constant? It's not "c=1" (except in informal shorthand), it's "c = 1 Planck speed unit". It seems like all sorts of things would "break" very quickly when you start dropping units. Maybe someone has a link that goes into more depth on why all of this is okay?
raattgift 9y ago
The value of c is arbitrary and depends on one's choice of coordinates. However the presence of c in the line element of a spacetime like ours is mandatory. There is, however, some conceptual value in setting c to some value other than 1, which I'll return to in the last paragraph below. When using a pseudo-Riemannian manifold one uses a metric signature where (keeping it simple, cf. "metric signature" on wikipedia) coordinates on orthogonal dimensions of can take one sign, or the opposite sign and a constant multiplier. In the Lorentzian case there will be one dimension taking one sign, and one or more taking the other; the choice of whether the solitary dimension takes a + or - sign is a matter of convention or preference. Conventionally the solitary dimension also takes the constant multiplier and is called the timelike dimension. A Lorentzian signature (minuses, 1) or (1, plusses) guarantees that one can describe paths through the manifold as null, timelike, or spacelike, and this gives one a causal structure. Our universe can be well represented by a Lorentzian manifold (3, 1) or (1, 3), and this has been tested to exquisite precision. It does not tell us the value of the constant c, but we can determine that from tests of causal relations, the boundaries of which will be null. Alternatively, a massless pointlike object will always travel on null geodesics. One runs into c being set to unity in systems geometrized units in relativity often; it's very handy to have mark off coordinates as e.g. "-seconds" vs "c seconds" (-,+,+,+ aka (1,3)) as long as one doesn't mess up one's dimensional analyses (which is unfortunately easy). As a concrete example, the line element for (1,3) flat spacetime using Cartesian coordinates is dS^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2. Compare the formula for Euclidean distance in the (Euclidean flat) plane between points p = (p1,p2) and q = (q1,q2) for (x,y) coordinates: d(p,q) = sqrt((q1-p1)^2 + (q2-p2)^2), which we could rewrite as dx = q1-p1, dy = q2-p2, ds = sqrt(dx^2 + dy^2) or ds^2 = dx^2 + dy^2. In Euclidean 3-space, we add another axis: ds^2 = dx^2 + dy^2 + dz^2. In Lorentzian 4-spacetime, we have to change the sign, so ds^2 = dx^2 + dy^2 + dz^2 - c^2dt^2. Note that we are not restricted to use any particular unit of distance or system of units; dx could be in metres, miles, astronomical units, light-years, gigaparsecs or practically anything else, while dt could be in seconds or fortnights or any other handy unit of time. When setting c to unity, we do need to choose appropriate units for dx (and dy and dz ...) vs dt. In SI units, that's seconds and light-seconds, but we could use another system if we wanted. Notably we aren't restricted to Cartesian coordinates, however if we were to change to some other system of coordinates (e.g. polar ones) the line element would need to reflect that. For example, in the Lorentzian 4-spacetime case, we would write ds^2 = dr^2 + r^2 * dtheta^2 + r^2sin^2(theta)dphi^2 -c^2dt^2. Finally, when using SI units (for example), one can see very clearly that a path taken by an object moving much slower than the speed of light is totally dominated by the amount of time between starting point and finishing point, because the value of c is large. Using the (+,-,-,-) metric signature [ds^2 = c^2dt^2 - dx^2 - dy^2 - dz^2], a large c helps make it clear that lightlike paths through spacetime are shorter, and purely timelike paths (where dx=0, dy=0, dz=0, dt != 0) have extremized length, since we don't subtract anything from cdt. This is the root of the explanation of the twin paradox in flat spacetime: the twin moving quickly compared to the speed of light takes a shorter path between together1 = (x1,y1,z1,t1) and together2 = (x2,y2,z2,t1) than the twin moving slowly compared to the speed of light. In the extreme, twin A holds x=const,y=const,z=const, whereas twin B's x coordinate is only equal to twin A's at the start and end of the journey. This holds up under any system of coordinates; we could consider r=const vs changing r in spherical coordinates, for instance.
BeetleB 9y ago
I won't profess knowledge of particle physics, but in EM systems where they set c=1, c is not dimensionless.

mikeash 9y ago

Consider the more mundane example of heat and work. The standard unit of heat is the calorie, which is the heat needed to increase the temperature of one gram of water by one degree celsius under standard conditions. The standard unit of work is the joule, which is the work needed to exert a force of one newton over a distance of one meter. James Joule discovered that heat and work can be related by: h = kw Where k is a constant equal to 4.184 calories/joule. But really, this isn't a relationship between heat and work. This is a statement of equivalence. The constant of 4.184cal/J just tells you how the two units differ. They both describe the same property. Really, h = w. The version with k is only needed if you insist on using the traditional, different units on the left and right sides. The same is true of e = mc^2. This does not tell you how two different fundamental properties are related, it merely tells you how two units, traditionally considered separate, actually measure the same underlying property. Really, e = m. The c^2 is only needed if you insist on using J on one side and kg on the other. That is to say, you absolutely can have a kilogram of energy. If you have an object with a mass of one kilogram, you do have a kilogram of energy. Physicists and engineers don't typically use that unit for energy, but there's no reason they couldn't, and likewise no reason they couldn't use joules for mass.

throwaway287391 9y ago
Thank you for the nice explanation, despite my somewhat rude and uninformed posts! I had some trouble accepting it, but the heat/work example made it "click" for me, once my brain got past its initial stubborn reaction of "but of course heat and work are both energy, whereas mass is clearly not!" So any time two physical quantities are related by a constant factor -- even if said constant factor has what we currently think of as "units" (and even particularly weird units like squared speed) -- they are fundamentally equivalent. I'm still mulling this over, but it's a pretty fascinating mental revelation for me.
- mikeash 9y ago
  Nice. I thought the heat/work thing was an interesting angle on it and I'm glad it worked. Edit: I also want to say that I didn't find the comment I replied to rude in any way. It's direct and straightforward, but that's pretty common around here.