Short post this week.<\/p>\n\n\n\n
I want to talk about a simple concept that is extremely useful in physics and other fields of applied science. Especially those fields working with equations to calculate anything that has a physical meaning to it, like a speed, a voltage, a weight..<\/p>\n\n\n\n
As a physics student, our teachers would usually spend a few minutes talking about how it is extremely important to check dimensions. Back then, I was listening and not really thinking how that would help me concretely to find the answer to the problem they were giving us. I would always be puzzled by why they were talking about this, and then never, ever, using it themselves in class or during the exercises. It seemed like a gimmick that I could not understand.<\/p>\n\n\n\n
As a physics tutor for bachelor students, I made sure to always explain why checking the dimensions is the most important habit to have. I even thought to myself: Why didn’t our teacher spent more time on that? It’s so useful!<\/p>\n\n\n\n
I try to make posts related to what I live through. This week, I had to review some technical text and low and behold, there was an equation. Simple equation. To be honest, this is an equation I don’t know by heart, it’s an equation I always derive from other base equations. It’s a habit I have to avoid remembering too many redundant things that I can simply find again (prove again) in less than a minute. That way I found out I make less mistakes.<\/p>\n\n\n\n
Anyway, that equation is used to find the optical path length for a given phase difference. The equation I saw was this one:<\/p>\n\n\n\n
OPL = \\dfrac{2\\pi \\times \\Delta\\phi}{\\lambda}<\/span><\/p>\n\n\n\n It’s wrong. And I could immediately see that in 5 seconds. How? by checking the dimensions of the equation.<\/p>\n\n\n\n First things to know are the dimensions of the components of the equation. OPL is a distance, we can note that [m]. 2\\pi<\/span> is representing an angle in this case specifically, we can note that [rad]. \\lambda<\/span> is the wavelength, it is a distance, we can also note that [m]. \\Delta \\phi<\/span> is the phase, it has a dimension of an angle, we note that [rad]…<\/p>\n\n\n\n Let’s replace the components of the equation by the dimensions…<\/p>\n\n\n\n [m] \\neq \\dfrac{[rad]\\times[rad]}{[m]}<\/span><\/p>\n\n\n\n As you can notice, the right side of the equation has a strange dimension: angle squared divided by a length, while the left side of the equation is a simple distance… Not equal at all, the equation is false. Do I need to know the details of the equation to know it is false? No. I can tell this equation is false just by this simple trick that cost me 5-10 seconds depending on the complexity of the equation…<\/p>\n\n\n\n Now, this technique is helpful to know if an equation is false, but it certainly doesn’t tell you if the equation is right…<\/p>\n\n\n\n Consider:<\/p>\n\n\n\n OPL = \\dfrac{2\\pi \\times \\lambda}{\\Delta\\phi}<\/span><\/p>\n\n\n\n Make the dimension analysis:<\/p>\n\n\n\n [m] = \\dfrac{[rad]\\times[m]}{[rad]}<\/span><\/p>\n\n\n\n You divide angles by angles, end up with a ratio and the only dimensions left is a distance in both sides of the equation. Yet, the equation is false.<\/p>\n\n\n\n So checking the dimensions is not a proof by itself that an equation is right. It is only a proof that an equation is false or that it could be right…. Nuance!<\/p>\n\n\n\n For the curious, the real, correct equation is the following:<\/p>\n\n\n\n OPL = \\dfrac{\\Delta\\phi\\times\\lambda}{2\\pi}<\/span><\/p>\n\n\n\n The dimension analysis would give the same result as the previous equation, ending in a length in both sides of the equation.<\/p>\n\n\n\n Tips and tricks<\/strong>: how to remember this equation? This equation is basically a ratio between phase and length. The main idea is that when your wave (light in this instance) travels a given distance of 1 wavelength, the phase will have turned 360\u00b0 making a full circle. Because we usually work in radian we use 2\\pi<\/span> instead.<\/p>\n\n\n\n What happens when the wave travels a given distance, here OPL, then the phase will have turn a given amount \\Delta \\phi<\/span>. Because this is a simple proportionality relation between OPL and \\Delta \\phi<\/span> and we know the proportion for a OPL of 1 wavelength, we can easily find the relation:<\/p>\n\n\n\n \\dfrac{OPL}{\\Delta\\phi} = \\dfrac{\\lambda}{2\\pi}<\/span><\/p>\n\n\n\n which is the equation I recreate every time I need to use it instead of remembering by heart.<\/p>\n","protected":false},"excerpt":{"rendered":" Short post this week. I want to talk about a simple concept that is extremely useful in physics and other fields of applied science. Especially those fields working with equations to calculate anything that has a physical meaning to it, like a speed, a voltage, a weight.. As a physics student, our teachers would usually spend a few minutes talking about how it is extremely important to check dimensions. Back then, I was listening and not really thinking how that would help me concretely to find the answer to the problem they were giving us. I would always be puzzled … <\/p>","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_coblocks_attr":"","_coblocks_dimensions":"","_coblocks_responsive_height":"","_coblocks_accordion_ie_support":"","_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"footnotes":""},"categories":[7,11],"tags":[50,49,46],"class_list":["post-300","post","type-post","status-publish","format-standard","hentry","category-optics","category-physics","tag-dimensions","tag-equations","tag-physics"],"aioseo_notices":[],"aioseo_head":"\n\t\t\n\t\n\t\n\t\n\t\n\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t